Percent Composition Of Chemical Compounds MCQ Quiz

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%

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.

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%. 

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

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.