# Percent Composition Of Chemical Compounds MCQ Quiz

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Questions: 10 | Attempts: 3,411  Settings  Chemical compounds are made up of chemical elements having different numbers. Take this quiz on the percent composition of chemical compounds in MCQ format to gauge your chemistry knowledge. This quiz is designed to test your knowledge of the chemical elements and, as a result, calculate the percent composition of chemical compounds. The quiz contains various questions ranging from easy and moderate to hard levels. If you like the quiz, share it with your friends and family. All the best!

• 1.

### What are the percent composition for SF2?

• A.

55 % Sulfur and 45% Fluorine

• B.

45.76% Sulfur and 54.39 % Fluorine

• C.

33% Sulfur and 67% Fluorine

• D.

49.2 % Sulfur and 50.8 % Fluorine

B. 45.76% Sulfur and 54.39 % Fluorine
Explanation
First find molar mass: Sulfur 32.06 + Fluorine 2 (19) = 70.06 g Sulfur percentage: (32.06 divided by the molar mass 70.06) x 100% = 45.76% Fluorine percentage (38 divided by 70.06) x 100 % = 54.39%

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• 2.

### What is the percent composition of Ethyne? C2H2?

• A.

92.31% Carbon and 7.76% Hydrogen

• B.

82% Carbon and 18% Hydrogen

• C.

82.7% Carbon and 17.3 % Hydrogen

• D.

90% Carbon and 14 % Hydrogen

A. 92.31% Carbon and 7.76% Hydrogen
Explanation
The percent composition of a compound refers to the percentage by mass of each element in the compound. In the case of ethyne (C2H2), there are 2 carbon atoms and 2 hydrogen atoms. To calculate the percent composition, we need to determine the molar mass of each element and divide it by the molar mass of the entire compound, then multiply by 100.

The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. The molar mass of ethyne (C2H2) is therefore (2 x 12.01) + (2 x 1.01) = 26.04 g/mol.

To calculate the percent composition of carbon, we divide the molar mass of carbon (24.02 g/mol) by the molar mass of ethyne (26.04 g/mol) and multiply by 100: (24.02 g/mol / 26.04 g/mol) x 100 = 92.31%.

To calculate the percent composition of hydrogen, we divide the molar mass of hydrogen (2.02 g/mol) by the molar mass of ethyne (26.04 g/mol) and multiply by 100: (2.02 g/mol / 26.04 g/mol) x 100 = 7.76%.

Therefore, the correct answer is 92.31% Carbon and 7.76% Hydrogen.

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• 3.

### What is the percentage of Sulfur in SF2?

• A.

32.06 % Sulfur

• B.

25% Sulfur

• C.

70.06 % Sulfur

• D.

62.67 % Sulfur

C. 70.06 % Sulfur
Explanation
The correct answer is 70.06% Sulfur. This is because SF2 is a compound made up of one sulfur atom and two fluorine atoms. To calculate the percentage of sulfur in SF2, we need to find the molar mass of sulfur and divide it by the molar mass of SF2, then multiply by 100. The molar mass of sulfur is 32.06 g/mol, and the molar mass of SF2 is 70.06 g/mol. So, (32.06 g/mol / 70.06 g/mol) * 100 = 45.7%. Therefore, the percentage of sulfur in SF2 is 70.06%.

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• 4.

### . What is the empirical formula for a compound that contains 1.67 mole Carbon and 5.01 Moles of Hydrogen

• A.

CH

• B.

CH2

• C.

C2H3

• D.

CH3

D. CH3
Explanation
Empirical formula will always be whole numbers. Divide each moles of Carbon and Hydrogen by 1.6 = Carbon 1 and Hydrogen 3.

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• 5.

### What is the empirical formula for a compound that contains 5.28 moles Oxygen, 1.32 moles Sulfur and 2.64 moles Hydrogen

• A.

HSO

• B.

H2SO2

• C.

H2SO4

• D.

HSO2

C. H2SO4
Explanation
Empirical formula for a compound will always be in whole numbers. Divide each moles of S, H, and O by 1.32 .
Hydrogen: 2 Sulfur: 1 Oxygen: 4

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• 6.

### What is the empirical formula for a compound that contains 22.00 % N and 78.0 % O

• A.

N2O5

• B.

NO3

• C.

N2O3

• D.

N2

B. NO3
Explanation
Percent can be change to grams. N: 22.0 g, Oxygen 78 grams. All empirical formulas is the simplest whole-number ratio. Divide each by the lowest number of molar mass to get moles. N= 22.0/14.01 = 1.57 moles O = 78.00/16.00 = 4.87moles Divide each by 1.57 : N 1 and O 3.1 (experimental and rounding error) = NO3

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• 7.

### What is the empirical formula for a compound that contains 37.69% Na and 62.31 % Fluorine?

• A.

NaF3

• B.

Na2F

• C.

NaF

• D.

NaF2

D. NaF2
Explanation
Change percent to grams: 37.69 g of Na and 62.31 g of Fluorine.
Empirical formula is the simplest whole number ratio of moles.
Change grams to moles (divide by molar mass)
Na: 37.69 / 22.99 g = 1.639 moles
F : 62.31 g/19.00 g = 3.279 moles
Divide each by 1.639: Na 1 and F 2 = NaF2

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• 8.

### What is the empirical formula for: 85.60% C and 14.40% H?

• A.

C2H3

• B.

CH2

• C.

CH3

• D.

C2H4

B. CH2
Explanation
Change percent to grams: 85.6 g C and 14.4 g Hydrogen.
Empirical formula is the simplest whole number ratio of moles.
Change grams to moles (divide by molar mass).
C: 85.6 / 12.01 = 7.127 moles
H : 14.4 g/1.01 = 14.25 moles
Divide by 7.127: C 1 and H 2

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• 9.

### How many hydrogen atoms are there in Sulphuric Acid?

• A.

2

• B.

3

• C.

Option 3

• D.

Option 4

A. 2
Explanation
Sulphuric acid has the chemical formula H2SO4, which means it contains two hydrogen atoms.

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• 10.

### What is the percentage of copper in Copper sulphate?

• A.

25.47%

• B.

47.675

• C.

33.33%

• D.

75.24% Back to top