Stoichiometry and Formulas

Question 1

From which of the following formulas can molecular mass be derived? Molecular Empirical Both

Answer 1

Molecular only

Question 2

What is the empirical formula of C6H12O6

Answer 2

CH2O

Question 3

A researcher wants to determine the empirical formula for a 75-g sample containing 30 g of carbon, 5 g of hydrogen, and 40 g of oxygen. What is the empirical formula of this sample?

Answer 3

CH2O

Question 4

What is the molecular formula of a sample weighing 56 g that is 48 g of carbon and 8 g of hydrogen?

Answer 4

We can’t actually find the molecular formula from the information given, but we can calculate the empirical formula. To do so, we need to convert the grams of each substance into moles using the atomic mass for each element. 48 g of carbon (~ 12 g/mol) corresponds with 4 moles and 8 g of hydrogen (~ 1 g/mol) is 8 moles. Thus, our ratio is 4 C : 8 H or 1:2. From here, though, we can’t determine what the molecular formula is; we just know it must satisfy the ratio of 1 carbon to 2 hydrogens.

Question 5

What is a possible molecular formula of a molecule with a molar mass of 120 g/mol and an empirical formula of CH2O?

Answer 5

C4H8O4

Question 6

An unknown compound is found to contain 80% carbon and 20% hydrogen by mass. What is this compound’s empirical formula?

Answer 6

CH3

Question 7

A particular compound has the empirical formula C2H6ON. What is an approximation of the percent composition by mass of nitrogen in this compound?

Answer 7

about 25%

Question 8

How many grams of CO2 can be produced when combusting 88 grams of propane?

Answer 8

264 grams

Question 9

The preparation of acetic acid from ethanol can be represented as follows:

C2H5OH (aq) + O2 (g) → CH3COOH (aq) + H2O (l)

If 92 grams of ethanol are reacted with excess oxygen, how many total moles of acetic acid are produced?

Answer 9

2 mols …Since ethanol has a molar mass of about 46 g/mol and there is a 1:1 ratio of ethanol to acetic acid in this balanced equation, we can do the following calculations:

92 g EtOH × (1 mol EtOH / 46 g EtOH) × (1 mol acetic acid / 1 mol EtOH) = 2 moles.

Question 10

Methane bubbling from the ocean floor mixes with water, as represented by the following chemical equation:

CH4 (g) + H2O (l) → CO (g) + 3H2 (g)

In a biology experiment, 30 mg of methane and 18 mg of water are mixed in a closed container.

True or false: The limiting reagent in this experiment is H2O.

Answer 10

This statement is true. To calculate the limiting reagent, we must determine how much methane is required to react with 18 mg of water. First, we need to convert grams to moles. 0.018 g H2O × 1 mol/18 g = 0.001 mol water, which requires 0.001 mol CH4, or 0.016 g, to fully react. Since we actually have 0.03 g of CH4, which is more than we need, water must be the limiting reagent.

Question 11

Methane and water react according to the equation CH4 (g) + H2O (l) → CO (g) + 3 H2 (g). If 60 mg methane and 36 mg water are mixed in a closed container, what is the theoretical yield of carbon monoxide?

Answer 11

56 mg

We have 0.06 g methane × 1 mol methane / 16 g, or 0.00375 mol methane. We also have 0.036 g water × 1 mol water / 18 g , or 0.002 mol water. Since these two compounds react in a 1:1 ratio, water is the limiting reagent, as we have less of it. Next, we need to find how much carbon monoxide forms from the complete reaction of 0.002 mol water. Again, a 1:1 ratio means that 0.002 mol CO are produced. Since CO has a molar mass of 28 g/mol, this is equivalent to 0.056 g, or 56 mg, of carbon monoxide.

Question 12

After completing the experiment from the previous question (in which 60 mg of methane and 36 mg of water reacted according to the equation CH4 (g) + H2O (l) → CO (g) + 3 H2 (g)), a student finds she produced 42 mg of carbon monoxide. What was the percent yield in her experiment?

Answer 12

According to the calculations from the last question, the theoretical yield of CO is 56 mg. Since the student’s actual yield was 42 mg, the percent yield can be calculated as 42 mg / 56 mg = 0.75 = 75%.

Question 13

What is the empirical formula for an unknown compound if a 92 g sample contains roughly 52% carbon, 15% hydrogen, and 33% nitrogen by mass?

Answer 13

C2H7N.

To solve this question, the first step is to convert the mass percentages into mass values and then convert that ratio into a molar ratio. In a 92 g sample, 52% is roughly 48 g, 15% is roughly 14 g, and 33% is roughly 30 g. Using the atomic mass of the elements in question, there are 4 moles of carbon, 14 moles of hydrogen, and approximately 2 moles of nitrogen. Thus, the ratio is 4:14:2 or 2:7:1 (C2H7N)

Question 14

What is the molecular formula of an unknown sample that is 92% carbon and 8% hydrogen by mass?

Answer 14

Cannot be determined

Question 15

What is the molecular formula of a molecule with an empirical formula C3H7N and a molar mass of 174 g/mol?

Answer 15

C9H21N3

To determine the molecular formula from the empirical formula and the molar mass, the most straightforward approach is to calculate the molar mass of the empirical formula and then divide the given molar mass by that value to determine how many times the empirical formula should be repeated. In this case, 3 carbons, 7 hydrogens, and 1 nitrogen combine to give a mass of 57 g/mol. 171 g/mol of our unknown / 57 g/mol = 3. So there should be 3 “copies” of the empirical formula, giving 9 carbons, 21 hydrogens, and 3 nitrogens, for a molecular formula of C9H21N3

Question 16

An unknown compound is found to contain roughly 42% carbon, 10% hydrogen, and 48% nitrogen by mass. What is this compound’s empirical formula?

Answer 16

With percent composition questions, if not given a molar mass, it’s useful to imagine a 100 g sample of the compound and then use that to determine the ratio of elements. In a 100 g sample of this unknown, there are 42 g of carbon, 10 g of hydrogen, and 48 g of nitrogen. Given carbon’s 12 g/mol atomic mass, there are 42/12 or 3.5 moles of carbon. In 10 g of hydrogen, there are 10 moles. In 48 g of nitrogen, there are about 3.5 moles of nitrogen. Thus the ratio is 3.5:10:3.5, or about 1:3:1, giving an empirical formula of CH3N.

Question 17

Copper(II) sulfide and nitric acid are reacted, yielding copper(II) nitrate, sulfur, water, and nitrogen monoxide. When six moles of copper(II) sulfide and sixteen moles of nitric acid are mixed in a reaction vessel, which is the limiting reagent?

Answer 17

First, the balanced equation described by the question must be written: 3 CuS + 8 HNO3 → 3 Cu(NO3)2 + 3 S + 4 H2O + 2 NO. Since the 6:16 molar ratio of copper sulfide to nitric acid is equivalent to the 3:8 ratio in the balanced equation, the two reagents are present in stoichiometric quantities. Neither is limiting; both will be exhausted if the reaction runs to completion.

Question 18

A student prepares to complete a lab report covering decomposition reactions. The first reaction studied is the decomposition of ammonium dichromate:

(NH4)2Cr2O7 (s) → Cr2O3 (aq) + H2O (l) + N2 (g)

Theoretically, how many grams of nitrogen gas should be obtained from the decomposition of 750 grams of ammonium dichromate? (Note: Ammonium dichromate’s molar mass is approximately 250 g/mol.)

Answer 18

84 grams

Question 19

Another student completing the same lab performs the same decomposition reaction:

(NH4)2Cr2O7 (s) → Cr2O3 (aq) + H2O (l) + N2 (g)

The student attempts to decompose 500 grams of ammonium dichromate, but only measures 12 grams of N2 gas produced. What is this student’s percent yield?
(Note: Ammonium dichromate’s molar mass is approximately 250 g/mol.)

Answer 19

21%

Since the molar mass of ammonium dichromate is about 250 g, 500 g represents 2 moles of the compound. Complete decomposition of this amount should produce 2 mol of N2, or 56 g. To find the percent yield, take the actual yield and divide it by the theoretical yield:
Percent yield = 12/56 = 0.21 = 21%.

Question 20

How many grams of hydrogen gas are required to completely react with 32 g of oxygen to form hydrogen peroxide?

Answer 20

The formation reaction for hydrogen peroxide is:

H2 (g) + O2 (g) → H2O2 (l)

32 g O2 x (1 mol/32 g) x (1 H2/1 O2) x 2 g/mol = 2 g H2

Question 21

If 70 g of nitrogen gas is reacted with 10 g of hydrogen gas, what mass of ammonia gas will be produced?

Answer 21

57g