Chem 107 Worksheet: Empirical and Molecular Formulas Name: For all calculated answers, show all work, round to the proper number of significant figures and box your final answer Student ID 1. A compound consists of 26.641 g of aluminum and 8.8944 g of carbon. Based on these amounts, what would be its empirical formula and compound name? 2. A compound is found to consist of the following composition: 31.42 % sulfur, 31.35 % compound? oxygen, and 37.23 % fluorine by weight. What is the empirical formula for this 3. The 38.25-gram product of a newly tried synthesis process was analyzed to consist of 74.01% rubidium, 5.20% carbon, and 20.78% oxygen. a. Indicate the mass breakdown of each element in the product b. Calculate the empirical formula of the product and a name for the compound 4. If 68.412 grams of cobalt metal is mixed with excess sulfur and heated strongly, a 124.250-gram sulfidle compound is produced. Calculate the empirical formula of the sulfide. 5. A compound with the empirical formula C2140 was found in subsequent experiments to have a molar mass of 220.26 g/mol. What would be the molecular formula of the compound? 6. A 17.28-gram sample of a new compound is analyzed to consist of 70.58% Hg, 12.47% Cl and 16.95% O. a) Give the masses of elements in the sample; b) calculate an empirical formula of the compoundi, c) If the new compound is determined to have a molar mass of 568 g/mol, find the molecular formula.

1.

Empirical formula of the compound is: Al₄C₃

Name of the compound is: Aluminum carbide

2. Empirical formula of the compound is: SO₂F₂

3.

a.

Mass of Rb in the product = 28.31 g

Mass of C in the product = 1.989 g

Mass of O in the product = 7.948 g

b.

Empirical formula of the product is: Rb₂CO₃

Name of the compound: Rubidium carbonate

4. Empirical formula of the sulfide is: Co₂S₃

5. Molecular formula of the compound is: C₁₀H₂₀O₅

6. 

a. 

Mass of Hg in the sample = 12.20 g

Mass of Cl in the sample = 2.155 g

Mass of O in the sample = 2.929 g

b. Empirical formula of the compound is: HgClO₃

c. Molecular formula of the compound is: Hg₂Cl₂O₆

Step-by-step explanation

1.

The compound contains 26.641 g Al and 8.8944 g C

Empirical formula and name of the compound = ?

Mass of Al = 26.641 g

Molar mass of Al = 26.981539 g/mol

Moles = mass/molar mass

Moles of Al = 26.641 g/26.981539 g/mol = 0.9874 mol

Mass of C = 8.8944 g

Molar mass of C = 12.0107 g/mol

Moles of C = 8.8944 g/12.0107 g/mol = 0.7405 mol

Mole ratio of Al:C = 0.9874 mol:0.7405 mol = 1.333*3:1*3 = 4:3

Empirical formula of the compound is: Al₄C₃

Name of the compound is: Aluminum carbide

2. 

The compound contains 31.42% S, 31.35% O, and 37.23% F

Empirical formula of the compound = ?

We assume the compound contains 100 g, so the percentage compositions are assumed to be in grams

Mass of S = 31.42 g

Molar mass of S = 32.065 g/mol

Moles of S = 31.42 g/32.065 g/mol = 0.9799 mol

Mass of O = 31.35 g

Molar mass of O = 16 g/mol

Moles of O = 31.35 g/16 g/mol = 1.959 mol

Mass of F = 37.23 g

Molar mass of F = 18.998403 g/mol

Moles of F = 37.23 g/18.998403 g/mol = 1.960 mol

Mole ratio of S:O:F = 0.9799 mol:1.959 mol:1.960 mol = 1:2:2

Empirical formula of the compound is: SO₂F₂

3.

a.

Mass of the product = 38.25 g

Composition of the product is: 74.01% Rb, 5.20% C, and 20.78% O

Mass of Rb in the product = 38.25 g*74.01/100 = 28.31 g

Mass of C in the product = 38.25 g*5.20/100 = 1.989 g

Mass of O in the product = 38.25 g*20.78/100 = 7.948 g

b.

Empirical formula of the product and the name of the compound = ?

Mass of Rb  = 28.31 g

Molar mass of Rb = 85.4678 g/mol

Moles of Rb = 28.31 g/85.4678 g/mol = 0.3312 mol

Mass of C = 1.989 g

Molar mass of C = 12 g/mol

Moles of C = 1.989 g/12g/mol = 0.1658 mol

Mass of O = 7.948 g

Molar mass of O = 16 g/mol

Moles of O = 7.948 g/16 g/mol = 0.4968 mol

Mole ratio of Rb:C:O = 0.3312 mol:0.1658 mol:0.4968 mol = 2:1:3

Empirical formula of the product is: Rb₂CO₃

Name of the compound: Rubidium carbonate

4. 

Mass of the sulfide compound = 124.250 g

Mass of cobalt, Co = 68.412 g

Mass of S in the compound = Mass of the sulfide compound - Mass of cobalt, Co

Mass of S in the compound = 124.250 g - 68.412 g = 55.838 g

Molar mass of Co = 58.933195 g/mol

Moles of Co = 68.412 g/58.933195 g/mol = 1.161 mol

Molar mass of S = 32.065 g/mol

Moles of S = 55.838 g/32.065 g/mol = 1.741 mol

Mole ratio of Co:S = 1.161 mol:1.741 mol = 1:1.5 = 2:3

Empirical formula of the sulfide is: Co₂S₃

5. 

Empirical formula of the compound: C₂H₄O

Molar mass of the compound = 220.26 g/mol

Molecular formula of the compound = ?

C₂H₄O contains 2 C atoms, 4 H atoms, and 1 O atom

Atomic mass of C = 12 g/mol

Atomic mass of H = 1 g/mol

Atomic mass of O = 16 g/mol

(C₂H₄O)_n = 220.26

(2*12 + 4*1 + 1*16)n = 220.26

44n = 220.26

n = 220.26/44 = 5

Molecular formula of the compound is given by (C₂H₄O)₅ = C₁₀H₂₀O₅

So,

Molecular formula of the compound is: C₁₀H₂₀O₅

6. 

a. 

Mass of the new compound = 17.28 g

The compound contains 70.58% Hg, 12.47% Cl, and 16.95% O

Mass of each element = ?

Mass of Hg in the sample = 17.28 g*70.58/100 = 12.20 g

Mass of Cl in the sample = 17.28 g*12.47/100 = 2.155 g

Mass of O in the sample = 17.28 g*16.95/100 = 2.929 g

b. 

Mass of Hg in the sample = 12.20 g

Molar mass of Hg = 200.59 g/mol

Moles of Hg = 12.20 g/200.59 g/mol = 0.06082 mol

Mass of Cl in the sample = 2.155 g

Molar mass of Cl = 35.453 g/mol

Moles of Cl = 2.155 g/35.453 g/mol = 0.06078 mol

Mass of O in the sample = 2.929 g

Molar mass of O = 16 g/mol

Moles of O = 2.929 g/16 g/mol = 0.1831 mol

Mole ratio of Hg:Cl:O = 0.06082 mol:0.06078 mol:0.1831 mol = 1:1:3

Empirical formula of the compound is: HgClO₃

c. 

Molar mass of the compound = 568 g/mol

Molecular formula of the compound = ?

(HgClO₃)_n = 568

(1*200.59 + 1*35.453 + 3*16)n = 568

284n = 568

n = 568/284 = 2

Molecular formula of the compound is: Hg₂Cl₂O₆