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


Atomic Weights

Now that you have some idea of what is in an atom, let's look at the weight of atoms. Each atom has a definite and characteristic weight. This weight provides a very convenient way to state the amount of substance required for a chemical reaction.

In this chapter we discuss how the weights of atoms were determined experimentally. You will encounter for the first time a very formidable number — 602,200,000,000,000,000,000,000 (6.022 × 1023) — called Avogadro's number. The number becomes very important in later chapters, so be sure you understand its significance!

OBJECTIVES

After completing this chapter, you will be able to

· recognize and apply or illustrate: isotope, atomic weight, atomic mass unit (amu), mass spectrograph, gram atomic weight, and Avogadro's number;

· explain the fractional atomic weights listed in the periodic table;

· calculate the number of atoms in a given weight of an element and vice versa;

· calculate the approximate atomic weight of an element when given the abundance and approximate mass of its isotopes;

· distinguish between atomic weight expressed in amu and gram atomic weight.

Image Let's review a bit. The notation Image indicates a neutral atom of chlorine.

1. What is its atomic number? _________

2. What is its mass number? _________

3. How many protons does it have? _________

4. How many electrons? _________

5. How many neutrons? _________

Answer: (a) 17; (b) 35; (c) 17; (d) 17; (e) 18

Image Different atoms of the same element can have different numbers of neutrons and, therefore, different mass numbers. Here is another neutral chlorine atom: Image.

1. What is its mass number? _________

2. How many protons does it have? _________

3. How many neutrons? _________

Answer: (a) 37; (b) 17; (c) 20

Image Since neutrons and protons combine to make up the mass number, two atoms of the same element can have different mass numbers. Chlorine can exist as Image and as Image. The only difference between these atoms of chlorine is that Image contains two more neutrons than Image.

Antimony can exist as Image and Image. The only difference between these atoms of antimony is that Image contains two more ___________________ than Image.

Answer: neutrons

Image Image has a greater mass than Image because of the two extra neutrons. Which of the following atoms of antimony has the greater mass, Image or Image? _______________

Answer: Image (because it has two more neutrons)

Image Atoms of the same element having different masses are called isotopes. Elements as found in nature are usually mixtures of two or more isotopes. The atom Image is one isotope of the element antimony; Image is another isotope of antimony. The main difference between two isotopes of the same element is the number of (protons, neutrons, electrons) _______.

Answer: neutrons

Image Isotopes exist for every known element. The isotopes of the element neon were first discovered by two English scientists, J. J. Thomson and F. W. Aston. Thomson and Aston continued in their work to discover other isotopes through inventing the mass spectrograph (also called the mass spectrometer).

In the mass spectrograph, atoms of different masses of the same element (mixtures of isotopes) are charged (no longer neutral) and accelerated by an electron beam toward a target, such as a photographic plate. A strong magnetic field bends the paths of the charged atoms. Atoms of greater mass have their paths bent to a lesser degree than atoms of lighter mass.

In the diagram of the mass spectrograph (pictured here) where do the lighter atoms strike, point A or point B? _______

Image

Answer: point A (because the path of the lighter atoms is bent to a greater degree)

Image An analogy to the mass spectrograph would be to roll a bowling ball and a basketball at the same speed at a target while a stiff crosswind is blowing. The bowling ball is considerably heavier than a basketball. Look at the diagram below. Which ball would strike the target at point B? _______

Image

Answer: the basketball (The basketball is lighter; therefore, its path is more readily changed by the crosswind.)

Image In the rolling balls analogy to the spectrograph, the bowling ball and basketball are analogous to isotopes of different mass. The strong crosswind is analogous to the _______________________. (Refer to the diagram of the spectrograph, if necessary.)

Answer: magnetic field (or magnet)

Image Thomson and Aston invented an instrument that detects the presence and characteristics of isotopes. What is this instrument called? _____________________

Answer: mass spectrograph (or mass spectrometer)

Image The atomic weights of the elements are listed in the periodic table. The atomic weight of sodium, for example, is listed as 22.990. The atomic weight listed for sodium is actually the atomic weight of a mixture of isotopes, Image and Image. The proportion of these isotopes is generally constant wherever sodium is found.

The atomic weight of an element is the average weight of a mixture of two or more ________.

Image

Answer: isotopes

Image The periodic table lists the atomic weight of Al as _____________________.

Answer: 26.982

Image Atomic weights are based on the carbon 12 scale. That is, carbon 12 or Image, the most abundant isotope of carbon, is used as the standard unit in measuring atomic weights. By international standard, one atom of the Image isotope has an atomic weight of exactly 12 atomic mass units, abbreviated amu. The atomic weight of the Image isotope is exactly ________ amu.

Answer: 12

Image All atomic weights can be expressed in atomic mass units. By international agreement, 12 amu would equal the mass of a single Image atom. One amu is equal to what fraction of a single Image atom? ________

Answer: Image the mass of a single Image atom

Image Although the Image isotope weighs exactly 12.000 amu by definition, the atomic weight of C as listed on the periodic table is 12.011 amu. The atomic weight of carbon as listed on the periodic table is greater than that of the Image isotope. Why? ______________

Answer: The atomic weight of an element is the average weight of a mixture of two or more isotopes.

Image While the element carbon as found in nature is made up largely of the Image isotope (98.9%), a small quantity of Image isotope (1.1%) is mixed uniformly as part of the element. The Image isotope has an atomic weight of 12.000 amu. The Image isotope has an atomic weight of 13.003 amu. The resultant average atomic weight would be slightly (heavier, lighter) ________ than 12.000 amu.

Answer: heavier (In fact, the periodic table lists the atomic weight of carbon as 12.011 amu.)

Image The atomic weights on the periodic table are the average atomic weights of the isotopic mixtures in the element. We can determine the average atomic weight of an element if we know the approximate mass of each isotope and the proportion of each isotope within the element.

Here are the steps for calculating the average atomic weight of the element carbon. Look at the table below as you read through the steps.

Multiply the mass of the Image isotope by its decimal proportion (12.000 × 0.989).

Multiply the mass of the Image isotope by its decimal proportion (13.003 × 0.011).

Add the results to find the average atomic weight of the element C.

Element

Isotope

Mass of isotope


Proportion in element


Mass × proportion



Sum

C

Image

12.000

×

0.989

=

11.868

+

=



Image

13.003

×

0.011

=

 0.143




Now you do the final calculation step, adding the two results. Fill in the blank in the “Sum” column. Round off this answer and all others to the nearest hundredth (two places to the right of the decimal) unless otherwise indicated.

Answer: The calculated atomic weight of C is 12.011, rounded off to 12.01 amu.

Image Now calculate the atomic weight for fluorine.

Element

Isotope

Mass of isotope


Proportion in element


Mass × proportion



Sum

F

Image

19.000

×

99.7% or 0.997

=


+

=



Image

18.000

×

0.3% or 0.003

=





Answer:

  Mass × proportion



  Sum

18.943

+

=

18.997

0.054




The calculated atomic weight of F is rounded off to 19.00 amu.

Image Sodium has two isotopes, Image and Image. The isotope Image has an atomic mass of approximately 23.000 amu, and its proportion in the element is 99.2%. The isotope Image has a mass of approximately 22.000 amu and a proportion within the element of 0.8%. Determine the atomic weight of sodium. (Remember, 0.8% = 0.008 and 99.2% = 0.992.) Use a separate sheet of paper to set up a table as we have done in the past few frames. Here again are the column headings you will need.

Element

Isotope

Mass ofisotope

Proportionin element

  Mass ×  proportion

Sum

Answer:

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Sum

Na

Image

23.000

×

0.992

=

22.816

+

=

22.992


Image

22.000

×

0.008

=

0.176




The calculated atomic weight of Na is 22.99 amu.

Image The element cobalt has an isotope Image that has an approximate mass of 60.00 and constitutes 48.0% of the element. Another isotope, Image, has an approximate mass of 58.00 and constitutes 52.0% of the element. Calculate the atomic weight of Co using the given data. Use a separate sheet of paper to set up a table of calculations.

Answer:

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Sum

Co

Image

60.00

×

0.480

=

28.80

+

=

58.96


Image

58.00

×

0.520

=

30.16




The calculated atomic weight of Co is 58.96 amu.

Image The percentage proportions of all the isotopes within an element must add up to a total of ________ %. The decimal proportions of all the isotopes within an element must add up to ________ (Hint: See above example, 0.480 + 0.520.)

Answer: 100; 1

Image We've been calculating atomic weight given the mass and proportion of isotopes. We can also determine the proportion of each individual isotope within an element if we know the atomic weight of the element. Set up a table like the ones you have been using and fill in the information given below. Use letters such as A and B to represent unknown proportions.

The element Cr, which has an overall atomic weight of 51.996 amu, has two isotopes: Image with atomic mass of 52.000 amu and Image with an atomic mass of 51.000 amu. Just fill in the “givens” and “unknowns” for now; don't try to solve the problem yet.

Answer:

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Sum

Cr

Image

52.000

×

A

=

52.000 × A

+

=

51.996


Image

51.000

×

B

=

51.000 × B




Image To solve this equation with two unknowns, you must form a second equation showing the relationship between A and B. You have already learned the answer to the following question in frame 20. Add the decimal proportions. A + B = _____________

Answer: 1

Image Modify the equation in frame 22 so that just B remains on the left side of the equation. B = _______

Answer: 1 − A

Image Here is the equation you need for calculating the sum of the isotopes' mass × proportion: (52.000A) + (51.000B) = 51.996. Substitute the expression (1 − A) for B in the equation _______

Answer: Image

Image Solve the equation derived in frame 24 to determine the value of A (the proportion of Image in an average mixture of chromium) to the nearest thousandth. Use a separate sheet of paper for your calculations.

Answer:

Image

Image Since B = 1 − A, what is the value of B (nearest thousandth)? B = _______

Answer:

Image

Image Here is a similar problem. In the next few frames, you will determine the proportion of the Image isotope in an average mixture of chlorine, which is made up of both Image and Image. Here is our table of givens and unknowns.

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Atomic weight

Cl

Image

34.97

×

A

=

34.97 × A

+

=

35.45


Image

36.97

×

B

=

36.97 × B




The values of A and B added together must equal _______.

Answer: 1

Image Since A + B = 1, then B = _______

Answer: 1 − A

Image The proportion 1 − A has been substituted for B in the following table.

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Atomic weight

Cl

Image

34.97

×

A

=

34.97 × A

+

=

35.45


Image

36.97

×

(1 − A)

=

36.97 × (1 − A)




Using the table above, determine the proportion of Image within the element chlorine (find the value of A). Use a separate sheet of paper for your set of equations.

Answer:

Image

Image Determine the proportion of Image in Cl (find B).

Answer: The proportion of Image in Cl has been given the value of B in the table.

Image

Image Neon has two isotopes: Image and Image. The approximate mass of Image is 22.000 amu and the mass of Image is approximately 20.000 amu. The atomic weight of neon is 20.179 amu. Determine the proportion (to the nearest thousandth) of the Image isotope within the element. (Let the proportion of Image within Ne be equal to B.) Use a separate sheet of paper for your table of calculations and set of equations.

Answer:

Element

Isotope

Mass of isotope


Proportion in element


  Mass × proportion



Atomic weight

Ne

Image

22.000

×

A =

=

(22.000 × A)

+

=

20.179


Image

20.000

×

B =

=

(20.000 × B)




Image

Image

Image What is the proportion (to the nearest thousandth) of the other isotope, Image, within the element Ne? _______

Answer: The proportion of Image is equal to the value of B in the table.

Image

Image So far, we have considered all atomic weights in terms of atomic mass units (amu).

An atomic weight expressed in amu represents the average weight of how many atom(s) of an element? _______

Carbon has an atomic weight of 12.011 amu, which represents the average weight of how many atom(s) of carbon? _______

Answer: (a) one (An atomic weight expressed in amu represents the average weight of one single atom of an element.); (b) one

GRAM ATOMIC WEIGHT

Image Since it is impossible to measure the weight of one atom with a laboratory balance, another unit for expressing atomic weight must be used. Atomic weight can be expressed in grams as well as amu. An atomic weight expressed in grams (called a gram atomic weight) contains 6.022 × 1023 atoms. This number, called Avogadro's number, will be encountered often in this book.

1. If the atomic weight of carbon is expressed as 12.011 amu, it represents the average weight of how many atom(s)? _______

2. If the atomic weight of carbon is expressed as 12.011 grams, it represents the average weight of how many atom(s)? _______

Answer: (a) one; (b) 6.022 × 1023

Image Using information in frame 34, answer the following question. One gram is how many times heavier than 1 amu? _______

Answer: 6.022 × 1023 (One gram is equivalent to 6.022 × 1023 amu.)

Image One ton is equivalent to 2000 pounds. One pound represents Image of a ton. 6.022 × 1023 amu is equivalent to 1 gram. One amu represents what fraction of a gram? ______

Answer:

Image

Image Avogadro's number (6.022 × 1023) is written in exponential notation (sometimes called scientific notation). It actually represents a very large number: 602,200,000,000,000,000,000,000. Exponential notation will be used throughout this book, as in most other chemistry texts. You may already be familiar with exponential notation and with multiplying and dividing numbers with exponents. If so, you may skip to frame 38. If you need a quick refresher of exponential notation, we have summarized the basic rules in the examples below.

Image

The exponent indicates the number of places that the decimal must be moved. The number 645,000 has the decimal moved five places to the left Image. The result is 6.45 × 105.

We can also use exponential notation as in the examples below.

Image

Image

In these two cases, we moved the decimal to the right, with the result that the exponent is negative.

Image (Move the decimal five places to the right to arrive at the notation 4 × 10−5.)

Image (Move the decimal two places to the right to arrive at the notation 7.3 × 10−2.)

For 0.073, we could have moved the decimal three places to the right, with a result of 73 × 10−3. We could also move the decimal one place to the right, with a result of 0.73 × 10−1. Normally, in chemistry calculations we write a single digit of value 1 through 9 to the left of the decimal.

To multiply two numbers with exponential notation, multiply the decimal portions of the numbers and add the exponents.

Image

To divide two numbers with exponential notation, divide the decimal portions of the numbers and subtract the exponent in the denominator from the exponent in the numerator.

Image

Try these problems.

1. 42,000 = 4.2 × _____________

2. 0.00465 = 4.65 × _____________

3. (7.0 × 10−5) × (4.0 × 108) = _____________

4. Image

5. Image

Answer:

1. 4.2 × 104

2. 4.65 × 10−3

3. (7.0 × 4.0) × 10(−5 + 8) = 28 × 103 = 2.8 × 104

4. Image

5. Image

Image Look at the following examples of fractions that are converted to exponential notation. Then fill in the correct answer for the equivalence of amu to grams by using the same process of eliminating the fraction and expressing the equivalence in proper exponential notation.

· 1 pound equals Image ton

· 1 pound equals Image ton

· 1 pound equals Image ton

· 1 pound equals 0.5 × 10−3 ton

· 1 pound equals 5.0 × 10−4 ton

· 1 amu equals Image grams

· 1 amu equals _________________________________

Answer: Image

Image You should memorize Avogadro's number (6.022 × 1023) and its meaning.

1. How many atoms are in an atomic weight expressed in grams?_______________

2. One gram is equivalent to the weight of how many amu?__________________

Answer: (a) 6.022 × 1023 atoms per gram atomic weight; (b) 6.022 × 1023 amu per gram

Image The average atomic weight of neon is 20.180 according to the periodic table (rounded off to the nearest thousandth).

1. The average weight of one neon atom is how many amu? _______

2. The average weight of 6.022 × 1023 neon atoms is how many grams? ______________

Answer: (a) 20.180; (b) 20.180

Image There are 6.022 × 1023 atoms in 1 gram atomic weight (atomic weight expressed in grams). The element neon has a gram atomic weight of 20.180 g. How many atoms are contained in 10.090 grams of neon? _______ (Hint: 10.090 grams of neon is half of the gram atomic weight of neon.)

Answer: Since 10.090 grams is half of a gram atomic weight, the number of atoms is half of 6.022 × 1023 atoms, or 3.011 × 1023 atoms.

UNIT FACTOR ANALYSIS (FACTOR LABEL ANALYSIS/DIMENSIONAL ANALYSIS)

Image If the problem in frame 41 had been more difficult, we would have used the unit factor method (also called factor label analysis and dimensional analysis in some texts) for solving problems, a mathematical procedure used in most chemistry textbooks for its convenience in calculations. We would have set up the problem in the manner shown below.

Image

All of the unit names except “atoms” cancel out during multiplication.

Image

Just the numbers and the name “atoms” are left after cancelation of unit names.

Image

The unit factor method involves multiplying the given value by one or more conversion factors. In this problem, we multiplied the 10.0895 grams of Ne by the conversion factors of

Image

The conversion factors come from definitions. For example, 1 gram atomic weight is equal to 6.022 × 1023 atoms, and 20.180 grams of Ne are equal to 1 gram atomic weight of Ne. The conversion factors are arranged so that the unit names will cancel out. They convert the units of the given values to those of the answer being sought and will give us the correct numerical answer.

Here is another example using the unit factor method. The necessary definitions are: 1 meter = 39.37 inches, and 1 yard = 36 inches.

The first conversion factor can be Image or Image.

The second conversion factor can be Image or Image.

Suppose we wish to determine the number of meters in 2.12 yards. We arrange the conversion factors so that the answer will be in meters and all other unit names will cancel.

Image

Use these conversion factors to determine the number of yards in 3.55 meters (nearest hundredth).

Answer: Image

Image The gram atomic weight of silicon is 28.09 grams. Using the unit factor method shown in frame 42, calculate the number of atoms in 2.00 grams of silicon (Si).

Answer:

Image

Image The atomic weight of silicon is 28.09. Determine the average weight (in amu) of one silicon atom. _______ amu

Answer: 28.09 (same as the atomic weight).

Image There is a big difference between an atomic weight expressed in amu and an atomic weight expressed in grams. You have already determined that one silicon atom weighs 28.09 amu. Now calculate the weight of one silicon atom in grams.

Image

Answer: 4.66 × 10−23

Image Determine the mass in grams of one boron (B) atom. (The gram atomic weight of boron is 10.81 grams.)

Answer:

Image

Image If one boron atom weighs 1.795 × 10−23 grams, what would 1,000,000 atoms of boron weigh? (1,000,000 = 1 × 106)

Answer: Image

Image The atomic weight of gold is 196.97. If you were offered an atomic weight of gold for only one dollar, would you buy it? Why or why not?


Answer: We wouldn't. The atomic weight of gold would be expressed as 196.97 amu, so you are buying one atom of gold for a dollar, definitely no bargain. (However, a gram atomic weight of gold weighs 196.97 grams, which would definitely be a bargain for one dollar.)

Image How much would a billion billion atoms (1 × 1018 atoms) of magnesium weigh? The gram atomic weight of magnesium is 24.305 grams. (Hint: Find the weight of one atom of magnesium and multiply the result by 1 × 1018.)

Answer: Here are two methods for solving this problem. The first method is one long step.

Image

The second method involves two shorter steps.

Image

What are the most important concepts in this chapter?

· The weights assigned to atoms are relative weights. That is, all atoms are compared to the weight of a single Image atom.

· All atoms of the same element do not have the same weight (elements exist as isotopes).

· The isotopes of an element are not present in nature in equal amounts.

· The atomic weight of an element may be determined experimentally.

· Atoms have very small weights. A chemist deals with large numbers of atoms (on the order of 6 × 1023) in the laboratory.

In Chapter 3 you will see that atomic weight is one of several periodic properties. You will encounter atomic weights throughout the remainder of this book, especially in Chapters 4 and 7, so you should have their meaning and significance clearly in mind before proceeding.

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. Calculate answers to the nearest hundredth unless otherwise indicated.

1. Isotopes of the element neon were first discovered by two English scientists, ______ and _____.

2. What is a mass spectrometer used for? _______

3. In a mass spectrometer's magnetic field which atoms are affected to a greater degree? _______ (lighter or heavier mass)

4. The atoms Image and Image are called _______ of chlorine.

5. How many protons, neutrons, and electrons are found in each of the chlorine isotopes in question 4?

6. How many neutrons are in the following carbon isotopes?

carbon-12 _______, carbon-13 ________, carbon-14 _______

7. What are the atomic weights of the following elements?

chromium (Cr) _______, potassium (K) _______________, aluminum (Al) _______

8. What are the atomic weights of the following elements?

osmium (Os) _______, calcium (Ca) ______________, gallium (Ga) _______

9. What elements have the following atomic weights?

4.00 _________

183.84 ________

244 _______

10. You are given 0.100 gram atomic weight of gold (Au). The atomic weight of gold is 196.97 grams to the nearest hundredth.

1. How many atoms would you have? _______

2. How many grams? _______

11. A sample of iron has a total of 9.77 × 1021 atoms of iron. Calculate the mass, in grams, of this sample.

12. Calculate the number of platinum atoms in a 15.5-gram sample of platinum. The atomic weight of platinum is 195.08 amu to the nearest hundredth.

13. The atomic weight of iron is 55.85 amu to the nearest hundredth and it has isotopes with approximate masses of 55.000 amu and 56.000 amu. What is the proportion of the 5626Fe isotope? (The other isotope is 5526Fe.)

14. The atomic weight of sodium is 22.99 amu to the nearest hundredth. A sample has isotopes with approximate masses of 20.998 and 23.991 amu. What is the proportion of the 24Na isotope? (The other isotope is 21Na.)

15. If you could cash it in, which would you rather have, 1 × 1040 amu of silver or 150 grams? (Hint: The problem is to determine which value has the greater number of atoms.)

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. J.J. Thomson and F. W. Aston (frame 6)

2. to detect the presence of isotopes of each element (frame 6)

3. lighter atoms are affected to a greater degree (frame 6)

4. isotopes (frames 5, 6)

5. Chlorine-35 has 17 protons, 18 neutrons, and 17 electrons. Chlorine-37 has 17 protons, 20 neutrons, and 17 electrons (frames 1—5).

6. Carbon-12 has six neutrons, carbon-13 has seven neutrons, and carbon-14 has eight neutrons (frames 1—5).

7. Cr = 51.996, K = 39.0983, Al = 26.98154 (frame 10)

8. Os = 190.23, Ca = 40.078, Ga = 69.723 (frame 10)

9. He (4.00), W (183.84), Pu (244) (frame 10)

10.

1. Image

2. Image

(frames 34, 39—43)

11. 0.906 g Fe (frames 42—49)

12. 4.78 × 1022 atoms Pt

13. Following the examples in the tables in frames 27-31, let A represent the proportion of 5626Fe isotope, and let 1 − A represent the proportion 5526Fe isotope.

Image

(frames 21—32)

14. Let A represent proportion: 24Na isotope, and let 1 − A represent proportion 21Na isotope.

Image

(frames 21—32)

15. Ag = 107.87 amu/atom, so our two equivalences are:

Image

You would probably rather cash in the 1 × 1040 amu.

(frames 47—49)

KILOGRAM'S CHANGE

Have you ever considered how important it is to have standards of measure? Without them how would we know how much mass an object possesses or how much time it takes for a chemical reaction to occur? There are many standards of measure for all kinds of measurements whether it be for mass, length, volume, time, etc.

Around the world the standard unit for mass is the kilogram. For more than 100 years the kilogram was defined by the mass of a platinum-iridium alloy that was housed at the International Bureau of Weights and Measures in Paris, France. The kilogram mass has served as the base unit of mass in the International System of Units (SI) from 1889 until present day.

ImageA 1-kilogram platinum-iridium alloy.

The international unit for mass, the kilogram, is used in chemistry and physics for a variety of applications. For example, if a chemist needs to know how many kilograms are in a sample of carbon s/he can obtain that through the atomic mass of carbon from the periodic table and then convert to kilograms.

Also, we can convert kilograms to another unit of measure. For instance, the mass of a 1983 penny is 2.5 × 10−3 kilograms, but we can convert this amount into grams.

Image

The kilogram is not only involved in converting between moles and kilograms, or kilograms to another unit of measure. The unit of mass used to express atomic and molecular weights is equal to one-twelfth of the mass of an atom of carbon-12, which is approximately 1.66 × 10−27 kilograms.

In addition to the kilogram there are 17 derived units that are defined in relation to the kilogram. In other words the kilogram is part of their unit of measure. Some of these include the Newton (N), Pascal (Pa), Joule (J), volt (V), Ohm (Ω), and Sievert (Sv). If we look at forces and Newton's second law of motion we see that force (F) = ma. Force is measured in Newtons, where m is the mass in kilograms and a is the acceleration of the object. One Newton is the equivalent of one kilogram multiplied by a meter per second squared.

Image

As you can imagine, how a kilogram is defined affects every derived unit mentioned. As of 20 May 2019 the International Committee for Weights and Measures approved a proposed redefinition of the kilogram. This redefinition, although small in its change, defines the kilogram in terms of the second and the meter. With the change in the kilogram's definition we can obtain even greater accuracy with the results obtained from a variety of experiments.