Question

Calculate the percent composition of these compounds: (Ethane:C2H6) and (Sodium hydrogen sulfate: NaHSO4).

Answer 1

To find the percent composition of ethane and sodium hydrogen sulfate, one must divide the mass of each element by the total molar mass of the compound and multiply by 100%. For ethane, the calculations yield 79.9% Carbon and 20.1% Hydrogen. For sodium hydrogen sulfate, the results are 19.15% Sodium, 0.84% Hydrogen, 26.72% Sulfur, and 53.29% Oxygen.

Step-by-step explanation:

Percent Composition Calculation

To calculate the percent composition of a compound, you need to divide the mass of each element in one mole of the compound by the molar mass of the compound and multiply by 100%.

Percent Composition of Ethane (C2H6)

The molar mass of ethane is 2(12.01 g/mol) + 6(1.008 g/mol) = 30.07 g/mol. The percent composition is calculated as follows:

• Carbon: (2(12.01)/30.07)*100% = 79.9%
• Hydrogen: (6(1.008)/30.07)*100% = 20.1%

Percent Composition of Sodium Hydrogen Sulfate (NaHSO4)

The molar mass of sodium hydrogen sulfate is 22.99 g/mol (Na) + 1.01 g/mol (H) + 32.07 g/mol (S) + 4(16.00 g/mol) (O) = 120.06 g/mol. The percent composition is calculated as:

• Sodium: (22.99/120.06)*100% = 19.15%
• Hydrogen: (1.01/120.06)*100% = 0.84%
• Sulfur: (32.07/120.06)*100% = 26.72%
• Oxygen: (4(16.00)/120.06)*100% = 53.29%

Answer 2

For the first compound,
Molar Mass of C=12.01x 2 =24.02
Molar Mass of H= 1.01x 6= 6.06
So molar mas of C2H6= (12.01x2)+(1.01x6) = 30.08
To get the percent composition, the formula is “(Total MM of C in molecule)/(Total MM of molecule)x 100”.
Here is the solution:
% of C= 24.02/30.08 x 100 = 79.9
% of H=6.06/30.08 x 100=20.1
To check if it is equal to 100%, you just have to add the percentages.
For the second compound,
Molar Mass of Na=22.99 x 1 =22.99
Molar Mass of H= 1.01x 1= 1.01
Molar Mass of S= 32.07x 1= 32.07
Molar Mass of O = 16.00 x4 = 64.00
So molar mas of NaHSO4= (22.99)+(1.01)+(32.07)+(16.00x4) = 120.07
To get the percent composition, the formula is “(Total MM of C in
molecule)/(Total MM of molecule)x 100”.
Here is the solution:
% of Na= 22.99/120.07 x 100 = 19.2
% of H= 1.01/120.07x 100 = 0.8
% of S= 32.07/120.07x 100 = 26.7
% of O= 64/120.07 x 100 = 53.3
To check if it is equal to 100%, you just have to add the percentages.