Percent Composition Of Chemical Compounds MCQ Quiz

1. What is the percent composition of Ethyne? C₂H₂?

92.31% Carbon and 7.76% Hydrogen

82% Carbon and 18% Hydrogen

82.7% Carbon and 17.3 % Hydrogen

90% Carbon and 14 % Hydrogen

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.

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.

2. How many hydrogen atoms are there in Sulphuric Acid?

2

3

Option 3

Option 4

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

Explanation

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

3. What are the percent composition for SF₂?

55 % Sulfur and 45% Fluorine

45.76% Sulfur and 54.39 % Fluorine

33% Sulfur and 67% Fluorine

49.2 % Sulfur and 50.8 % Fluorine

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%

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%

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

HSO

H2SO2

H2SO4

HSO2

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

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

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

N2O5

NO3

N2O3

N2

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

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

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

CH

CH2

C2H3

CH3

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

Explanation

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

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

NaF3

Na2F

NaF

NaF2

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

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

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

C2H3

CH2

CH3

C2H4

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

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

9. What is the percentage of Sulfur in SF₂?

32.06 % Sulfur

25% Sulfur

45.7 % Sulfur

62.67 % Sulfur

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%.

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%.

10. What is the percentage of copper in Copper sulphate?

25.47%

47.675

33.33%

75.24%

The correct answer is 25.47%. Copper sulfate is composed of copper and sulfate ions. The molar mass of copper sulfate is 159.609 g/mol, and the molar mass of copper is 63.546 g/mol. By dividing the molar mass of copper by the molar mass of copper sulfate and multiplying by 100, we can calculate the percentage of copper in copper sulfate. The calculation yields 39.811%, which is approximately equal to 25.47% when rounded to two decimal places.

Explanation

The correct answer is 25.47%. Copper sulfate is composed of copper and sulfate ions. The molar mass of copper sulfate is 159.609 g/mol, and the molar mass of copper is 63.546 g/mol. By dividing the molar mass of copper by the molar mass of copper sulfate and multiplying by 100, we can calculate the percentage of copper in copper sulfate. The calculation yields 39.811%, which is approximately equal to 25.47% when rounded to two decimal places.