Chemistry: A Self-Teaching Guide - Post R., Snyder C., Houk C.C. 2020

Molecular and Formula Weights

Since every molecule is made up of a definite and invariable number of component atoms, it follows that each kind of molecule has a definite and characteristic weight. You may ask, “How did anyone ever determine how many atoms of each kind are in a molecule of any pure substance?” We try to answer that question in this chapter. We may determine the molecular weight of a substance in a number of ways using instrumental techniques that are available today. Specific instrumental methods are not discussed here. Instead, we will rely upon the data that we get by analyzing substances and what you have learned so far in Chapters 1 through 3. We will depend upon the periodic table and the information it provides.

This chapter will teach you how to determine the molecular weight and percentage composition of a compound when given its molecular formula. You will also be able to reverse the process to determine the molecular formula of a compound given the percentage composition of the compound (or the weight of each element per molecule) and its molecular weight. The important thing to remember is that every molecule consists of a definite number of atoms in a fixed ratio expressed as small whole numbers.

OBJECTIVES

After completing this chapter, you will be able to

· explain the difference between a molecule of a compound and a molecule of an element;

· calculate the molecular weight of a compound when given its actual formula;

· explain the difference between an empirical formula and an actual formula;

· calculate the percentage by weight of each element in a compound when given its actual formula or empirical formula;

· determine the actual formula of a compound when given its empirical formula and molecular weight;

· calculate the actual formula of a compound when given its percentage composition and its molecular weight.

In Chapter 2 you learned that each atom has an average atomic weight expressed in atomic mass units (amu) and that atomic weights are included on the periodic table. You also learned that an atomic weight expressed in grams represents the weight of 6.022 × 10²³ atoms.

In Chapter 3 you learned that atoms combine to form compounds of two major categories, depending upon the kind of chemical bond involved. In this chapter you will learn how to determine the weights of compounds.

Just to review a bit, the atomic weight of a Cl atom is _______ amu.

What is the weight in grams of 6.022 × 10²³ atoms of Cl? _______

Answer: 35.453; 35.453

The chlorine gas molecule (Cl₂) is made up of two chlorine atoms. If one atom of chlorine weighs 35.45 amu (to the nearest hundredth), one molecule of Cl₂ should weigh _________. (Note: Calculate your answer to the nearest hundredth for this and further questions unless otherwise stated.)

Answer: 70.90 amu (35.45 × 2 = 70.90)

MOLECULAR WEIGHT

Carbon monoxide (CO) is a molecule made up of one atom of carbon and one atom of oxygen. If an atom of carbon weighs 12.01 amu and an atom of oxygen weighs 16.00 amu, a molecule of CO should weigh ___________________ (to the nearest hundredth).

Answer: 28.01 amu (12.01 + 16.00 = 28.01)

The molecular weight of a molecule is the sum of the combined atomic weights of the atoms within the molecule. If the atomic weight of H is 1.01 amu and the atomic weight of Cl is 35.45 amu, a molecular weight of HCl is ___________________ amu (to the nearest hundredth).

Answer: 36.46 (1.01 + 35.45 = 36.46)

In Chapter 3 you learned that a molecule of a compound consists of two or more different types of atoms that are chemically combined by one or more chemical bonds. The molecule carbon dioxide (CO₂) is made up of one atom of carbon and two atoms of oxygen. If the atomic weight of C is 12.01 amu and the atomic weight of O is 16.00 amu, the molecular weight of CO₂ is ___________________amu (to the nearest hundredth).

Answer: 44.01 (1 C + 2 0 = 12.01 + 32.00 = 44.01)

The term “molecule” can mean either a molecule of an element or a molecule of a compound.

A molecule of an element consists of one or more atoms of the same kind. Examples are H₂ (which exists as a diatomic molecule), Ar (which exists as a single argon atom), and S₈ (which exists as a molecule of eight sulfur atoms).

Since argon normally exists as a single atom, the single atom (Ar) can be called either the argon atom or the argon molecule. The atomic weight of Ar is 39.95 amu. The molecular weight of Ar is ___________________ amu.

Answer: 39.95 (The weight of one atom of Ar is the same as the weight of the Ar molecule because the Ar molecule is the Ar atom.)

A single atom molecule is an exception to the rule. Most molecules of either an element or a compound usually involve two or more atoms combined with covalent bonds.

The molecular weight of O₂ is _______ amu. The molecule O₂ is a molecule of a(n) (compound, element) _______.

Answer: 32.00 (twice the weight of a single O atom); element (since only one element, oxygen, is included in the molecule)

The atomic weight of S is 32.06 amu. The molecular weight of S₈ is _____ amu. S₈ is a molecule of an element because ___________________.

Answer: 256.48 (8 × 32.06 = 256.48); only one element (sulfur) is included in the molecule

The following example shows a simple procedure for finding the molecular weight of sucrose sugar (C₁₂H₂₂O₁₁).

1. The molecular weight of C₁₂H₂₂O₁₁ is _______ amu. (Hint: Add the results of Atoms × weight.)

2. Sucrose (C₁₂H₂₂O₁₁) is a molecule of a(n) (compound, element) _______.

Answer: 342.34; compound (because more than one element is included in the molecule: carbon, oxygen, and hydrogen)

Determine the molecular weight of ethyl alcohol (C₂H₅OH). From now on you should use the periodic table to determine the appropriate atomic weights (to the nearest hundredth). Below are the column headings we used for our table in frame 9. Use a separate sheet of paper and make a similar table to calculate the molecular weight of C₂H₅OH.

The molecular weight of C₂H₅OH is ___________________ amu.

Answer:

FORMULA WEIGHT

In Chapter 3 you learned that there were two general categories of compounds, depending upon whether the type of chemical bonding involved was covalent or ionic. Covalent bonding generally results in molecular compounds while ionic bonding results in ionic compounds. The term molecular weight, therefore, should properly refer only to molecular compounds and not to ionic compounds. Chemists have coined the term formula weight to refer to either molecular compounds or ionic compounds. The formula weight of a molecule is the same as the molecular weight. The formula weight can also refer to the weight of an ionic compound.

The weight of CO₂ (a covalent compound) is 44.01 amu. Is 44.01 amu the molecular weight, the formula weight, or either the molecular or the formula weight? ___________________

The weight of NaCl (an ionic compound) is 58.44 amu. Is 58.44 amu the molecular weight, the formula weight, or either the molecular or the formula weight? _______

Answer: either molecular or formula weight; formula weight

Determine the formula weight of copper sulfate, CuSO₄. Use a separate sheet of paper to set up a table similar to those in frames 9 and 10.

Answer:

Answer the following questions for vitamin C (C₆H₈O₆).

1. Calculate the formula weight of vitamin C. (Set up a table to calculate the amu.)

2. Vitamin C is a covalent compound. What is its molecular weight? ____ amu

Answer:

1.

2. 176.14 (Molecular weight is the same as the formula weight for a covalent compound.)

Carbon monoxide (CO) is a molecule with one atom of carbon and one atom of oxygen. The molecular weight of CO is 28.01 amu. The atomic weight of C is 12.01 amu. The atomic weight of O is 16.00 amu. Which atom (carbon or oxygen) accounts for more than half of the molecular weight of the CO molecule? ___________________

Answer: oxygen

PERCENTAGE COMPOSITION

To find the actual proportion by weight (percentage) of an element in a compound, the weight of the part must be divided by the weight of the whole. For example, a bolt and nut can be compared to a molecule with two atoms. If the nut weighed 100 amu and the bolt weighed 200 amu, the whole thing would weigh 100 + 200 = 300 amu.

The nut is a part of the whole. To find the percentage by weight that the nut occupies within the whole, divide the weight of the nut by the weight of the whole and multiply by 100. Here's how this percentage is calculated.

The percentage of the weight contributed by the nut is equal to 33.3% of the weight of the whole.

Assume that a bolt weighs 50 amu and a nut weighs 30 amu. What percentage by weight does the nut contribute in relation to the whole? The nut weighs ___________________% of the whole.

Answer:

A bolt has two nuts.

Each nut weighs 30 amu and the bolt weighs 50 amu. What percentage by weight do the nuts contribute in relation to the whole? The nuts weigh ___________________% of the whole.

Answer:

The carbon monoxide molecule can be considered analogous to the bolt and nut combination. The molecule consists of one carbon atom weighing 12.01 amu (the atomic weight of carbon) and one oxygen atom weighing 16.00 amu (the atomic weight of oxygen). The whole molecule weighs 12.01 + 16.00 = 28.01 amu (formula weight). The oxygen atom weighs ___________________% of the whole molecule.

Answer:

To find the percentage by weight of any element within a compound, divide the total atomic weight of that element by the formula weight of the compound and multiply the result by 100%.

The molecule carbon dioxide (CO₂) has a total formula weight of 44.01 amu and consists of one carbon atom with atomic weight of 12.01 and two oxygen atoms with combined atomic weights of _________ amu.

Answer: 32.00 (16.00 × 2 = 32.00 amu)

Using the information in frame 18, find the percentage by weight of oxygen within the compound carbon dioxide (CO₂). ________

Answer:

Now calculate the percentage by weight of carbon within carbon dioxide. ___________________

Answer: You could do this in two ways. Since you just found that carbon dioxide consists of 72.7% oxygen by weight, the rest must be carbon:

Or, if you did not already know the percentage of oxygen, then:

Calculate the percentage by weight of hydrogen (H) in water (H₂O). (Find the formula weight of H₂O first.)

% hydrogen = _____________________

Answer:

To find the percentage composition by weight:

Determine the percentage of carbon by weight in glucose (C₆H₁₂O₆).

% carbon = _______

Answer:

The formula weight of glucose (C₆H₁₂O₆):

In the preceding example (glucose, C₆H₁₂O₆), the percentage composition by weight of all elements (carbon, hydrogen, and oxygen) totaled together should equal _______%.

Answer: 100 (Since all the partial weights of all elements add up to the weight of the whole compound, all percentages also add up to their whole, which is 100%.)

EMPIRICAL FORMULA

A chemist often must determine the formula of a compound when only the percentage composition by weight is known. Assume a compound is composed of only two elements, carbon and hydrogen. By analysis, the compound is found to contain 92.2% carbon by weight. What is the percentage composition by weight of hydrogen in this compound? ________

Answer: 7.8% (All partial composition percentages must add up to 100%. Since there are only two elements in the compound, their percentages must add up to 100%. If carbon accounts for 92.2%, then hydrogen must account for the other 7.8%.)

A compound of undetermined formula contains 92.2% carbon by weight and 7.8% hydrogen by weight. This means that for every 100 amu of that compound, 92.2 amu would be contributed by carbon. How many amu would be contributed by hydrogen? ___________________

Answer: 7.8 amu

Knowing that 100 amu of a compound contains 92.2 amu of carbon and 7.8 amu of hydrogen gives us the relative weights of the elements in a compound. In order to determine a chemical formula, we need to know the relative number of atoms of each element in the compound. We need to know the ratio of carbon atoms to hydrogen atoms. A 100 amu sample of a compound contains 92.2 amu of carbon and 7.8 amu of hydrogen. We know that the atomic weight of carbon is 12.01 amu and represents the weight of one carbon atom. A weight of 92.2 amu represents the weight of how many carbon atoms (to the nearest tenth)? ___________

Answer:

A 100 amu sample of a compound contains 92.2 amu of carbon (representing 7.7 atoms of carbon) and 7.8 amu of hydrogen. How many atoms of hydrogen are contained in the sample (to the nearest tenth)? _______

Answer:

To arrive at a tentative formula for this unknown compound, we need to know the ratio of carbon atoms to hydrogen atoms within a single molecule. Since there are 7.7 atoms of hydrogen for 7.7 atoms of carbon, there must be how many atom(s) of hydrogen for each atom of carbon? _________

Answer: one

Since carbon and hydrogen are in a ratio of 1:1 in this unknown compound, the simplest formula is CH (or HC). The simplest possible chemical formula that represents the ratios of the atoms within an unknown molecule is called an empirical formula. The simple formula (CH) is called a(n) ______________.

Answer: empirical formula

An unknown compound with an empirical formula of CH could be any of several compounds. For example, both acetylene (C₂H₂) and benzene (C₆H₆) have an empirical formula of CH. The formulas for acetylene and benzene (C₂H₂ and C₆H₆) are called molecular formulas because they describe the actual number of atoms contained in each molecule.

1. A formula describing the simplest ratio between atoms in an unknown molecule is called a(n) _____________.

2. A formula describing the actual number of atoms contained in each molecule is called the _____________.

Answer: (a) empirical formula; (b) molecular formula

An empirical formula is sometimes the same as an actual molecular formula. A formula determined by percentage weight analysis is considered to be an empirical formula since it only represents the ratio of one atom to another. By weight analysis, a compound is found to consist of carbon and oxygen in a 1:1 ratio. A formula of CO is assigned to the compound and you suspect that the compound could be carbon monoxide but have no proof. The formula CO is in this case a(n) ___________ (empirical, molecular) formula.

Answer: empirical (If by simple weight analysis, the compound has a carbon and oxygen ratio of 1 : 1, then the assigned formula CO is empirical. If it is determined by additional testing that the compound is indeed carbon monoxide, then the formula CO would be a molecular formula.)

To determine an empirical formula if given percentage composition, follow these three steps.

1. Drop the percentage signs and replace them with amu. The result is the weight composition of a 100 amu sample.

2. Multiply each of the results of step 1 by the appropriate atoms per atomic weight conversion factor (as shown in frames 26 and 27). The result is the relative “number” of atoms in a 100 amu sample.

3. Divide each of the relative “number” of atoms obtained in step 2 by the smallest number to obtain the simplest ratio of atoms in whole numbers.

A compound is found to be 20% Ca and 80% Br by weight. Following step 1, a 100 amu sample of this compound would be composed of __________ amu of Ca and _________ amu of Br.

Answer: 20; 80

Using step 2 from frame 32, determine how many atoms (rounded to the nearest tenth) of the component elements are present in the 100 amu sample.

1. atoms of Ca = ______________

2. atoms of Br = ______________

Answer:

1.

2.

In this 100 amu sample of an unknown compound composed of 20 amu of Ca and 80 amu of Br are 0.5 atoms of Ca and 1.0 atoms of Br. However, 0.5 atoms is not a whole number. Using step 3 from frame 32 to obtain the simplest ratio of atoms in whole numbers, divide both 0.5 atoms of Ca and the 1.0 atoms of Br by the smallest number, which is 0.5. (Note: If the answers obtained in step 2 are whole numbers, step 3 is unnecessary.)

Answer: one; two

The simplest empirical formula derived from a ratio of one atom of Ca to two atoms of Br would be ______________.

Answer: CaBr₂

A compound is made up of 79.9% carbon and 20.1% hydrogen by weight. What is its empirical formula? (Apply the three steps from frame 32. Round to the nearest whole number.) ____________

Answer:

1. A sample of the unknown compound weighing 100 amu is made up of 79.9 amu of carbon and 20.1 amu of hydrogen (replacing “percentage” with “amu” to express weight composition).

2. The relative number of atoms in the 100 amu sample are as follows. A 79.9 amu weight of carbon contains 6.7 atoms of carbon.

A 20.1 amu weight of hydrogen contains 19.9 atoms of hydrogen.

3. We must divide the 6.7 atoms and the 19.9 atoms obtained in step 2 by the smallest number (6.7) to find the simplest whole number ratio of atoms.

4. The ratio of one atom of carbon to three atoms of hydrogen provides an empirical formula of CH₃ (or H₃C).

The same procedure can be used to determine the empirical formula of a compound with three or more elements. A sample of a compound is composed of 40% calcium (Ca), 12% carbon (C), and 48% oxygen (O) by weight. A 100 amu sample of this compound would be composed of _________ amu of Ca, ____________ amu of C, and ____________ amu of O.

Answer: 40; 12; 48

What is the empirical formula of the compound from the previous frame? (Write the formula in the order of the elements given, Ca, C, and O. Use the three-step procedure.)

Answer:

1. In a 100 amu sample, Ca = 40 amu, C = 12 amu, and O = 48 amu.

2. In that 100 amu sample there are:

3. Dividing by the smallest number given in step 2 in order to obtain a simple whole number ratio is not necessary in this case. We already have simple whole numbers.

The empirical formula must be CaCO₃ (elements listed from most metallic to least metallic).

Determine the empirical formula of a compound made up of 28.7% K, 1.5% H, 22.8% P, and 47.0% O. Write the formula in the order of the elements given (K, H, P, O). (In step 2, calculate to the nearest thousandth.) __________

Answer:

1. In a 100 amu sample, K = 28.7 amu, H = 1.5 amu, P = 22.8 amu, and O = 47.0 amu

2. In that 100 amu sample there are:

3. Dividing by the smallest number given in step 2 to find the simplest whole number ratio:

KH₂PO₄ is the empirical formula.

Determine the empirical formula of a compound composed of 15.15% K, 10.45% Al, 24.80% S, and 49.60% O. Write the formula in the order of the elements given (K, Al, S, O). (In step 2, calculate to the nearest thousandth.) ___________

Answer:

1. In a 100 amu sample, K = 15.15 amu, Al = 10.45 amu, S = 24.80 amu, and O = 49.60 amu.

2. In the 100 amu sample, there are:

3. Divide each of the above answers by the smallest answer to find the simplest whole number ratio.

KAlS₂O₈ is an acceptable empirical formula.

To review, one atom of chlorine (Cl) weighs ______ amu and 6.022 × 10²³ atoms of chlorine (Cl) weigh _______ grams.

Answer: 35.45; 35.45

The number 6.022 × 10²³ appears often in chemical calculations. The term mole means 6.022 × 10²³ chemical units and is much easier to write and remember than the number it represents.

· One mole of atoms means 6.022 × 10²³ atoms.

· One mole of electrons means 6.022 × 10²³ electrons.

· One mole of ions means 6.022 × 10²³ ions.

The current abbreviation for mole is mol. We will use the abbreviation in all calculations involving mole.

A gram atomic weight of chlorine (Cl) represents the weight of 1 _____ of chlorine atoms.

Answer: mole (mol) (6.022 × 10²³ atoms)

A mole of molecules represents how many molecules? ________

Answer: 6.022 × 10²³

The atomic weight in grams of calcium (Ca) is 40.08 grams. A mole of Ca atoms weighs how many grams? _______

Answer: 40.08

The element calcium (Ca) weighs _______ grams per gram atomic weight. The element calcium (Ca) weighs _______ grams per mole of atoms.

Answer: 40.08; 40.08 (A gram atomic weight is the weight of 1 mol of atoms.)

The mole is a very useful quantity in chemistry and is often encountered. You will begin using the mole in the next part of this chapter.

You have already used the conversion factor of .

You will now begin to use the conversion factor of .

For example, since 1 mol of chlorine (Cl) atoms weigh 35.45 grams, the conversion factor would be .

Instead of percentages, the composition of a compound may be given directly in weights. For example, a 5.00 gram sample of a compound (to the nearest hundredth gram) consists of only chlorine and phosphorus. If the 5.00 gram sample was known to contain 1.13 grams of phosphorus (P), how many grams of chlorine (Cl) would be present? _______

Answer: 3.87 (The 5.00 gram sample was of a compound with only two elements. Since one element made up 1.13 grams, the other element must account for the remainder: 5.00 − 1.13 = 3.87 grams.)

To find the empirical formula of the chlorine and phosphorus compound, first multiply the weights of the sample elements by the appropriate mole per gram atomic weight conversion factor (see frame 46). The result is the number of moles of each element present in the sample.

Use the same method to determine the number of moles of Cl atoms in the sample (rounded to the nearest thousandth).

Answer: 0.109

To determine an empirical formula for this sample, we must determine a whole number ratio of atoms. This can be done by simply dividing each mole quantity by the smallest number.

The resulting empirical formula is ________.

Answer: 3.0; PCl₃ or Cl₃P (One atom of P and three atoms of Cl. PCl₃ is best because the more metallic element is named first.)

The steps in determining an empirical formula when given a sample are similar to steps 2 and 3 encountered earlier (refer back to frame 32) when using percentage composition to determine empirical formula.

A. Determine the number of moles of atoms of each element by multiplying the sample portion of each element by its appropriate mole per gram atomic weight conversion factor.

B. Divide each answer from step A by the smallest answer to find the simplest whole number ratio of one atom to another.

A 10-gram sample of a compound is made up of 2.73 grams of carbon and 7.27 grams of oxygen. Complete step A below. (Round to the nearest thousandth.)

Answer: 0.454

Applying step B to the same sample:

Complete step B for oxygen (O) atoms. ___________

What is the resulting empirical formula of the sample? __________

Answer:

1.

2. CO₂

A 5.00-gram sample of a compound made up of calcium (Ca) and chlorine (Cl) is analyzed and contains 1.80 grams of Ca. The 5.00-gram sample contains how many grams of Cl? _________

Answer: 3.20 (5.00 − 1.80 = 3.20)

How many moles of atoms are represented by 1.80 grams of Ca and by 3.20 grams of Cl? (Calculate moles to the nearest thousandth.)

1. moles of Ca atoms = _________

2. moles of Cl atoms = _________

Answer:

1.

2.

Determine the simplest whole number ratio of Ca atoms to Cl atoms, and give the resulting empirical formula. _______________

Answer:

There is one atom of Ca for every two atoms of Cl, resulting in an empirical formula of CaCl₂.

A 10.00-gram sample of a compound is composed of 1.59 grams of boron (B) and 8.41 grams of fluorine (F). Determine the empirical formula of the compound. (Calculate step A to the nearest thousandth.) ____________

Answer:

A. Determine the number of moles of each element within the sample.

B. Determine the simplest whole number ratio of B atoms to F atoms.

The empirical formula is BF₃.

A 5.00-gram sample of a compound contains 3.74 grams of carbon (C) and is made up of only carbon (C) and hydrogen (H). Determine the empirical formula for this compound. (Hint: First find the weight of hydrogen in the sample. Calculate step B to the nearest thousandth.) ______________

Answer: Since the sample is 5.00 grams total and carbon (C) contributes 3.74 grams, the hydrogen (H) must weigh 1.26 grams (5.00 − 3.74 = 1.26).

A. Determine the number of moles of atoms of each element within the sample.

B. Determine the simplest whole number ratio of C atoms to H atoms.

The empirical formula is CH₄ (or H₄C).

An empirical formula weight can be expressed in either amu or grams. It is simply the combined atomic weights of the elements in the empirical formula. A sample of a compound has been found to have an empirical formula of CH (one carbon atom for every hydrogen atom). The empirical formula weight for CH is ________ amu.

Answer: 13.02 (C = 12.01 amu and H = 1.01 amu; thus, CH = 12.01 + 1.01 = 13.02 amu)

ACTUAL MOLECULAR FORMULA

The actual formula of a compound may be different from the empirical formula. For example, instead of CH, the actual formula could be C₂H₂ or C₆H₆. The empirical formula tells us only the ratio of one atom to another. In the case of the empirical formula CH, we know that the actual formula has one C atom for every H atom. After determining the empirical formula, a chemist must devise some other experimental means to determine the actual formula. This is often done by finding the formula weight of the actual compound. Suppose that, by experiment, it was determined that the actual formula weight of an unknown compound was double that of the empirical formula weight. The empirical formula is CH. The actual formula would be double that of CH. Which of the following formulas would you expect to be the actual formula: C₂H₂ or C₆H₆? ____________

Answer: C₂H₂

To determine the relationship between the empirical formula and the actual formula, divide the actual formula weight by the empirical formula weight. Suppose the actual formula weight has been found to be 78.11 amu. The empirical formula weight for CH is 13.02 amu.

1.

2. The actual formula is how many times larger than the empirical formula? ___________

3. If the empirical formula is CH, the actual formula would be _______.

Answer:

1. 6.0

2. 6 times larger (note that the actual weight is a whole number multiple of the empirical formula)

3. C₆H₆

A compound has been found to have an empirical formula of HO (one atom of hydrogen for every atom of oxygen). The actual formula weight has been found to be 34.02 amu. Determine the empirical formula weight of HO. (Remember, the empirical formula weight is the combined atomic weights of the elements in the empirical formula.)

1. The empirical formula weight of HO is ___________ amu.

2. What is the actual formula? __________

Answer:

1. 17.01(16.00 + 1.01 = 17.01 amu)

2. Divide the actual formula weight by the empirical formula weight to determine the relationship between the actual formula and the empirical formula.

The actual formula is two times as large as the empirical formula. Since the empirical formula is HO, the actual formula is H₂O₂.

A compound is found to have an empirical formula of CH₃. The empirical formula weight for CH₃ is _______ amu. (Use a separate sheet of paper to set up a table of calculation for this and the following frames.)

Answer:

The empirical formula is CH₃ and the empirical formula weight is 15.04 amu. The actual formula weight has been found to be 30.08 amu. What is the actual formula? __________

Answer: Divide the actual formula weight by the empirical formula weight:

The actual formula is two times as large as the empirical formula. Thus, the actual formula is C₂H₆.

A compound has an empirical formula of CH₂O. The actual formula weight has been found to be 180.18 amu.

1. Determine the empirical formula weight. _________

2. What is the actual formula? ____________________

Answer:

1.

2. Divide the actual formula weight by the empirical formula weight:

The actual formula is six times as large as the empirical formula. Thus, the actual formula is C₆H₁₂O₆.

A compound has an empirical formula of CO. The actual formula weight has been found to be 28.01 amu.

1. The empirical formula weight of CO is _____________.

2. What is the actual formula? ______________________

Answer:

1. 28.01 amu (16.00 + 12.01 = 28.01)

2. Since the actual formula weight and the empirical formula weight are the same, the empirical formula and the actual formula must also be the same.

The actual formula is CO.

Formula weights can also be expressed in grams. A compound has an empirical formula of NH₂. The actual formula weight has been found to be 32.06 grams.

1. Determine the empirical formula weight in grams. ______________

2. What is the actual formula? ______________

Answer:

1.

2. Divide the actual formula weight by the empirical formula weight.

The actual formula is twice as large as the empirical formula. Thus, the actual formula is N₂H₄.

A 10.00-gram sample of a compound is made up of 3.04 grams of nitrogen (N) and the remainder is oxygen (O). The actual formula weight of this compound is 92.02 grams. Determine the actual formula of this compound. (When determining the number of moles of each element, calculate to the nearest thousandth.)

Answer:

· First, calculate the amount of oxygen in the sample.

10.00 grams − 3.04 grams = 6.96 grams of oxygen

· Second, determine the number of moles of each element within the sample.

· Third, determine the simplest whole number ratio of N atoms to O atoms to find the empirical formula.

The empirical formula is NO₂.

· Fourth, determine the actual formula.

· Divide the actual formula weight by the empirical formula weight.

·

· The actual formula is twice as large as the empirical formula. Thus, the actual formula is N₂O₄.

What you have just learned is very important to a research chemist who is trying to determine the formula of a compound that has just been produced. The chemist can utilize a variety of instrumental and chemical processes to determine the percentage composition of the compound. In later chapters, you will learn that a chemist can also utilize several different ways to experimentally determine the formula weight of a compound. The combination of these methods permits the researcher to calculate the actual formula of the compound.

You will need to be able to determine formula weights in several of the remaining chapters of this book and in any chemistry course you take. It is important that you have a good understanding of Chapters 1 through 4 before continuing your study.

In Chapter 5 you will learn how chemists name the compounds they produce based upon an internationally accepted scheme.

SELF-TEST

This self-test is designed to show how well you have mastered this chapter's objectives. Correct answers and review instructions follow the test. Round answers to the nearest hundredth.

1. What is the formula weight of dihydrogen sulfide, H₂S, to the nearest hundredth? _______________

2. What is the percentage by weight of sulfur in dihydrogen sulfide? _______________

3. What is the formula weight of aspirin (acetylsalicylic acid), C₉H₈O₄? _______________

4. What is the percentage by weight of oxygen in aspirin? _______________

5. What is the formula weight of acetic acid, C₂H₄O₂? _______________

6. What is the percentage by weight of carbon in acetic acid? _______________

7. What is the actual formula of a compound that is 39.99% carbon, 6.67% hydrogen, and 53.34% oxygen if its actual formula weight is 150.15 amu? (Use a separate sheet of paper for your calculations.) _______________

8. How many moles of atoms are represented by 2.50 grams of Na and 4.44 grams of Br? (Calculate moles to the nearest hundredth.)

1. moles of Na atoms = _________________

2. moles of Br atoms = _________________

9. How many moles of atoms are represented by 0.35 grams of Mg and 0.98 grams of O? (Calculate moles to the nearest hundredth.)

1. moles of Mg atoms = _________________

2. moles of O atoms = _________________

10. What is the actual formula of a compound that is 92.26% carbon and 7.74% hydrogen, if its actual formula weight is 78.11 amu? (Use a separate sheet of paper for your calculations.)___________________

11. What is the actual formula of a compound that is 85.63% carbon and 14.37% hydrogen, if its actual formula weight is 56.11 amu? (Use a separate sheet of paper for your calculations.)

12. Lisinopril (C₂₁H₃₁N₃O₅) is a medication that is often prescribed to treat high blood pressure. The formula weight for lisinopril is 405.495 amu. Calculate the % carbon and % hydrogen for this compound.

13. Cadaverine (C₅H₁₄N₂) is a noxious compound produced from the putrefaction of animal tissue. The formula weight for cadaverine is 102.178 amu. Calculate the % N and % H of this compound.

14. Explain the difference between an empirical formula and the actual formula of a compound.___________________

15. How is a molecule of a compound different from a molecule of an element?___________________

ANSWERS

Compare your answers to the self-test with those given below. If you answer all questions correctly, you are ready to proceed to the next chapter. If you miss any, review the frames indicated in parentheses following the answers. If you miss several questions, you should probably reread the chapter carefully.

1. 34.08 amu (frames 4, 9, 13)

2. %sulfur = 32.06 amu of sulfur/34.08 amu of dihydrogen sulfide × 100 %  = 94.07 % S (frames 18—22)

3. 180.17 amu (frames 4, 9, 13)

4. (frames 18—22)

5. 60.06 amu (frames 4, 9, 13)

6. %carbon = [2 × (1 atom of carbon × 12.01 amu of carbon)/60.06 amu acetic acid] × 100 %  = 24.02/60.06 × 100% = 40%

7. Step 1:

o

o

o Step 3: The empirical formula is CH₂O

o Step 4:

Therefore, the actual formula is C₅H₁₀O₅. (frames 24—39, 56—59)

8.

1. (frame 52)

2. (frame 52)

9.

0. (frame 52)

1. (frame 52)

10.

o

o

o Step 3: The empirical formula is CH.

o Step 4:

Therefore, the actual formula is C₆H₆. (frames 24—39, 56—59)

11.

o Step 1:

o Step 2:

o Step 3: The empirical formula is CH₂.

o Step 4:

Therefore, the actual formula is C₄H₈. (frames 24—39, 56—59)

12.

(frames 15—23)

13.

(frames 15—23)

14. An empirical formula merely indicates the combining ratio of the atoms. The actual formula has the same atomic ratio as the empirical formula but includes the actual number of each atom present according to the measured formula weight. (frames 29—32)

15. A molecule of a compound consists of atoms of different elements, whereas a molecule of an element consists of atoms of the same element. (frame 6)