7.2.3: Practice Empirical and Molecular Formulas

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Percent Composition Problems

Exercise \(\PageIndex{1}\)

What is the percent composition of Co(NO₃)₂?

Hint: What is formula mass?

Co mass \(=1 \times 58.93 = 58.93\) amu

N mass \(=2 \times 14.01 = 28.02\) amu

O mass \(=6 \times 16.00 = 96.00\) amu

Total = 182.95 amu

Answer

Co mass percent =\(\dfrac{58.93amu}{182.95amu}\times 100\) % = 32.21 %

N mass percent =\(\dfrac{28.02amu}{182.95amu}\times 100\) % = 15.32 %

O mass percent =\(\dfrac{96.00amu}{182.95amu}\times 100\) % = 52.47 %

What is the percent composition of Ni₂(SO₄)₃?

Hint: What is formula mass?

Ni mass \(=2 \times 58.69 = 117.38\) amu

S mass \(=3 \times 32.07 = 96.21\) amu

O mass \(=12 \times 16.00 =192.00\) amu

Total = 405.59 amu

Answer

Ni mass percent =\(\dfrac{117.38amu}{405.59amu}\times 100\) % = 28.94 %

S mass percent =\(\dfrac{96.21amu}{405.59amu}\times 100\) % = 23.72 %

O mass percent =\(\dfrac{192.00amu}{405.59amu}\times 100\) % = 47.34 %

You analyze a sample and find that it contains 22.46 g of Fe and 42.76 g Cl. What is the percent composition of this sample?

Hint: What is total mass?

Your sample is 22.46 g + 42.76 g = 65.22 g

Answer

Fe mass percent =\(\dfrac{22.46g}{65.22g}\times 100\) % = 34.44 %

Cl mass percent =\(\dfrac{42.76g}{65.22g}\times 100\) % = 65.56 %

Empirical Formula from Percent Composition

Exercise \(\PageIndex{1}\)

What is the empirical formula of a compound that is 73.42 % Co and 26.58% O by mass?

Hint: Convert 73.42 g Co to moles Co (Co = 58.93 g/mol) and 26.58 g O to moles O (O = 16.00 g/mol).

Answer

At first, you get 1.246 mol Co and 1.661 mol O. Divide both of these by the smaller number (1.246)

This gives 1 mol Co and 1.33 mol O. Now multiply both of these by 3.

This gives 3 mol Co and 4 mol O.

Co₃O₄

What is the empirical formula of a compound that is 26.52 % Cr, 24.53 % S, and 48.95% O by mass?

Hint: Convert 26.52 g Cr to moles Cr (Cr = 52.00 g/mol), 24.53 g S to moles S (S = 32.07 g/mol), and 48.95 g O to moles O (O = 16.00 g/mol).

Answer

At first, you get 0.5100 mol Cr, 0.7649 mol S, and 3.059 mol O. Divide all of these by the smallest number (0.5100)

This gives 1 mol Cr, 1.5 mol S, and 6 mol O. Now multiply all of these by 2.

This gives 2 mol Cr, 3 mol S, and 12 mol O.

Cr₂S₃O₁₂ (this is actually the compound Cr₂(SO₄)₃)

Empirical and Molecular Formulas

Exercise \(\PageIndex{1}\)

For each formula below, give the empirical formula. Sometimes the formula given is the same as the empirical, sometimes it is different.

a) C₃H₆O₃

b) N₂O₄

c) Mg₃N₂

d) C₇H₁₄O₂

e) P₂O₅

f) C₄H₈N₂

Answer a and b: CH₂Oand NO₂

Answer c and d: Mg₃N₂and C₇H₁₄O₂

Answer e and f: P₂O₅ and C₂H₄N

What is the molecular formula for a compound with empirical formula of CH₂O and molecular mass of 150.15 amu?

Answer: C₅H₁₀O₅ (Molecular mass is five times as big as empirical formula mass.)

What is the molecular formula for a compound with empirical formula of CH₂NO₂ and molecular mass of 180.12 amu?

Answer: C₃H₆N₃O₆ (Molecular mass is three times as big as empirical formula mass.)

What is the molecular formula for a compound with a molecular mass of 86.18 amu that is found to be 83.62 % C and 16.38 % H by mass? (Must find empirical formula first using percent composition.)

Empirical Formula

C₃H₇

At first, you get 6.963 mol C and 16.22 mol H. Divide both of these by the smallest number (6.963)

This gives 1 mol C and 2.33 mol H. Now multiply both of these by 3.

This gives 3 mol C and 7 mol H.

Molecular Formula: C₆H₁₄(Molecular mass is two times as big as empirical formula mass.)

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