SAT Subject Test Chemistry
PART 2
REVIEW OF MAJOR TOPICS
CHAPTER 4
Chemical Formulas
Practice Exercises
Write the formula or name in questions 1 through 10:
1. AgCl _____________________________
2. CaSO_{4 }____________________________
3. Al_{2}(SO_{4})_{3 }__________________________
4. NH_{4}NO_{3 }_________________________
5. FeSO_{4 }____________________________
6. Potassium chromate__________________
7. Sodium fluoride ____________________
8. Magnesium sulfite ___________________
9. Copper(II) sulfate ___________________
10. Iron(III) chloride ___________________
Directions: The following set of lettered choices refers to the numbered questions immediately below it. For each numbered item, choose the one lettered choice that fits it best. Then fill in the corresponding oval on the answer sheet. Each choice in the set may be used once, more than once, or not at all.
Questions 1115
Use the following choices to indicate the oxidation state of the underlined symbol in the given formulas.
(A) +1
(B) +2
(C) +4
(D) +5
(E) +6
11. K_{2}CrO_{4}
12. Na_{2}S_{2}O_{3}
13. PO_{4}^{3}
14. CaCO_{3}
15. Mg(HCO_{3})_{2}
_________________________________________________
16. Find the percentage of sulfur in H_{2}SO_{4}.
17. What are the empirical formula and the true formula of a compound composed of 85.7% C and 14.3% H with a true formula mass of 42?
Directions:
Every question below contains two statements, I in the lefthand column and II in the righthand column. For each question, decide if statement I is true or false and whether statement II is true or false, and fill in the corresponding T or F ovals in the answer spaces. *Fill in oval CE only if statement II is a correct explanation of statement I.
I 
II 

18. In the formula of a compound, the algebraic sum of the oxidation numbers must be 0 
BECAUSE 
oxygen”s oxidation number in most compounds is −2. 
19. Fluorine is assigned an oxidation number of −1 in all compounds 
BECAUSE 
fluorine is the most electronegative element. 
20. Balanced equations have the same number of reactant atoms as the product atoms 
BECAUSE 
the conservation of matter must apply in all regular chemical equations. 
* Fill in oval CE only if II is a correct explanation of I.
21. Write the complete ionic equation for this reaction. Then write the net ionic equation.
3NaOH(aq) + Fe(NO_{3})_{3}(aq) → 3NaNO_{3}(aq) + Fe(OH)_{3}(s)
Complete ionic equation:
_________________________ →___________________________
Net ionic equation:
_________________________ →___________________________
Answers and Explanations
1. Silver chloride
2. Calcium sulfate
3. Aluminum sulfate
4. Ammonium nitrate
5. Iron(II) sulfate or ferrous sulfate
6. K_{2}CrO_{4}
7. NaF
8. MgSO_{3}
9. CuSO_{4}
10. FeCl_{3}
K^{+1} Cr^{x} O^{−2}
11. +6 because 2(+1) + 1(x) + 4(−2) = 0, x = +6
Na^{+1} S^{x} O^{−2}
12. +2 because 2(+1) + 2(x) + 3(2) = 0, x = +2
P^{x} O^{−2}
13. +5 because 1(x) + 4(−2) = −3, x = +5
Ca^{x} C^{+4} O^{−2}
14. (B) +2, because 1(x) + 1(+4) + 3(−2) = 0, x = +2
Mg^{+2} H^{x} C^{+4} O^{−2}
15. (A) +1, because 1(+2) + 2(x) + 2(+4) + 6(−2) = 0, x = 1
16. H_{2}SO_{4 }is composed of:
17. (C) = 12) 85.7% C H = 1) 14.3% H
7.14 14.3
To find the lowest ratio of the whole numbers:
7.14) 7.14 7.14) 14.3
1.0 2.0
The empirical formula is CH_{2}.
Since the formula mass is given as 42, the empirical formula CH_{2}, which represents a mass of 14, divides into 42 three times. Therefore, the true formula is C_{3}H_{6}.
18. (T, T) The sum of all the oxidation totals for a compound must equal 0, and oxygen in most compounds is assigned a −2 oxidation number, but this does not explain the I statement.
19. (T, T, CE) Statement II does explain statement I.
20. (T, T, CE) The Law of Conservation of Matter does require that there are equal numbers of respective atoms on both sides of a regular (not atomic) equation. Because the formula of each substance is determined by oxidation numbers, the equation is balanced by using coefficients.
21. From this formula equation:
3NaOH(aq) + Fe(NO_{3})_{3}(aq) → 3NaNO_{3}(aq) + Fe(OH)_{3}(s)
The complete ionic equation is:
3Na^{+}(aq) + 3OH^{−}(aq) + Fe^{3+}(aq) + 3NO_{3}^{−}(aq) → 3Na^{+}(aq) +
3NO_{3}^{−}(aq) + Fe(OH)_{3}(s)
The net ionic equation is:
Fe^{3+}(aq) + 3OH^{−}(aq) → Fe(OH)_{3}(s)