Lesson 6: Stoichiometry

Question 1

You have a substance that is composed of 20.8% Iron and 79.2% Chlorine by mass. What is the chemical formula of this compound (periodic table: http://www.chem.qmul.ac.uk/iupac/AtWt/table.gif)?

(A) FeCl3
(B) Fe2Cl3
(C) FeCl6
(D) Fe3Cl6

Answer 1

(C) FeCl6

Pretend you have 20.8 g of Fe. That would be 20.8 g / 55.845 g -> approx. .33 moles Fe
Pretend you have 79.2 g Cl. That would be 79.2 g / 35.45 g -> approx. 2 moles Cl.
So, for every 1 Fe atom, you must have 6 atoms of Cl, making the formula FeCl6.

Question 2

The _________ formula of glucose is C6H12O6 while the _________ formula of glucose is CH2O.

Fill in the blanks using the following options:

• Empirical
• Atomic
• Molecular
• Compound

Answer 2

The molecular formula of glucose is C6H12O6 while the empirical formula of glucose is CH2O.

Question 3

How would you find the molecular formula when given only the empirical formula?

Answer 3

You cannot find the molecular formula when only given the empirical formula.
You would also need to know the molecular mass to find the molecular formula from the empirical formula.

Question 4

How would you find the empirical formula when given only the molecular formula?

Answer 4

You can find the empirical formula by finding the ratio of atoms in the molecular formula and then dividing by their greatest common factor.

Question 5

1 gram is equal to:

(A) 6.022 ⋅ 10^23 molecules of Carbon
(B) 6.022 ⋅ 10^23 units of mass
(C) 6.022 ⋅ 10^23 atomic mass units
(D) 6.022 ⋅ 10^23 molecules of Helium

Answer 5

(C) 6.022 ⋅ 10^23 atomic mass units

1 gram is equal to 6.022 ⋅ 10^23 atomic mass units.

Question 6

What is the limiting reagent in the following reaction conditions with 7 moles A, 10 moles C, and 2 moles AC2, given the reaction:

1 mol A + 2 mol C -> 1 mol AC2

(A) A Only
(B) C Only
(C) AC2 Only
(D) A and C

Answer 6

(B) C Only

If 7 mol of A are available, and 1 mol of A is needed to form 1 mole of product, it could form up to 7 mol of AC2.

If 10 mol of C are available, and 2 mol C are needed to form 1 mol product, then up to 5 mol of AC2 could be produced. Thus, C is the limiting reagent.

Because AC2 is not a reagent, it can be excluded.

Question 7

To determine the number of moles in a given mass of a compound, you need to convert mass to moles. Which of the following terms do you divide the mass by to get moles?

(A) Atomic mass units
(B) Formula weight
(C) Avogadro’s number
(D) Percent composition

Answer 7

(B) Formula weight

The formula weight (i.e. molecular weight or molar mass) of a compound is the sum of the atomic weights of each atom in the compound.

Question 8

When the mass of a sample of a given compound is half of its formula weight, how many moles of the compound do you have?

(A) 1/4 of a mole
(B) 1/2 of a mole
(C) 1 mole
(D) 2 moles

Answer 8

(B) 1/2 of a mole

Recall that the formula weight of a compound is equal to the mass of one mole of the compound.

Question 9

You have 8.32 g of P4(s) and 3.22 g of Cl2. Considering the following unbalanced equation, how much PCl3 will be produced: P4 + Cl2 -> PCl3

(A) 4.12 g
(B) 1.82 g
(C) 0.27 g
(D) 15.78 g

Answer 9

(A) 4.12 g

The equation is balanced as follows: P4 + 6Cl2 -> 4PCl3
8.32 g P4 × (1 mol P4/[30.97×4] g P4) × 4 moles PCl3/mol P4 = approx. 0.25 mol PCl3 (actual: 0.269 mol PCl3)

3.22 g Cl2 × (1 mol Cl2/[35.45×2] g Cl2) × (4 moles PCl3/6 mol Cl2) = approx. 0.025 mol PCl3 (actual: 0.030 mol PCl3)

Thus, Cl2 is the limiting reagent.

0.03 mol PCl3 × 137.33 g/mol PCl3 = about 4g PCl3 (actual: 4.12 g PCl3)

Since Cl2 is the limiting reagent, and my answer will be close to 4g , making (A) 4.12 g the correct answer.

Question 10

Equivalents are related to limiting reagents, because using a different compound means a different mass needs to be used to have the same effect (whether in buffering, electrons to donate or protons to donate). How many grams of Magnesium would need to be used to donate as many electrons as 50g of Sodium (periodic table: http://www.chem.qmul.ac.uk/iupac/AtWt/table.gif)?

(A) 57.6 g
(B) 82.1 g
(C) 15.2 g
(D) 26.4 g

Answer 10

(D) 26.4 g

Notice that Magnesium could donate two electrons per atom, whereas Sodium could only donate one.

(50g Na)/(22.99g/mol Na) × (I mol e- / mol Na)= approx 2 mol e- (actual: 2.17 mol e-)

(2.17 mol e-) × (1 mol Mg/ 2 mol e-) × (24.3g Mg/ mol) = approx 25g Mg (actual 26.4g Mg)

Question 11

Normality of acids is also related to equivalents, comparing an acid’s potential proton donating to a monoprotic strong acid. What is the Normality of 5M H2S?

(A) 5 N
(B) 2.5 N
(C) 10 N
(D) 15 N

Answer 11

(C) 10 N

H2S is a diprotic acid, so you would multiply the molarity by 2 to get normality.