Respuesta :
Answer: The mass percent composition of nitrogen of the unknown compound is, 31.1 %
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yN_z+O_2\rightarrow CO_2+H_2O+NO_2[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and nitrogen respectively.
We are given:
Mass of [tex]CO_2=13.3g[/tex]
Mass of [tex]H_2O=9.52g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 13.3 g of carbon dioxide, [tex]\frac{12}{44}\times 13.3=3.63g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 9.52 g of water, [tex]\frac{2}{18}\times 9.52=1.06g[/tex] of hydrogen will be contained.
For calculating the mass of nitrogen:
Mass of nitrogen in the compound = (6.807) - (3.63 + 1.06) = 2.117 g
To calculate the percentage composition of nitrogen in sample, we use the equation:
[tex]\%\text{ composition of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100[/tex]
Putting values in above equation, we get:
[tex]\%\text{ composition of nitrogen}=\frac{2.117g}{6.807g}\times 100=31.1\%[/tex]
Hence, the mass percent composition of nitrogen of the unknown compound is, 31.1 %
Answer: 31.0%
Explanation:
mol C=13.3gCO2×1molCO244.009g×1molC1molCO2=0.3022 mol C
mol H=9.52gH2O×1molH2O18.015g×2molH1molH2O=1.057 mol H
Convert moles to grams.
0.3022 mol C×12.011 g C mol C=3.630 g
1.057 mol H×1.008 g H mol H=1.065 g
Subtract the grams of carbon and hydrogen from the total mass of the sample to obtain the mass of nitrogen in the sample.
6.807g−3.630 g−1.065 g=2.112 g
Divide by the total mass of the sample, 6.807g, and multiply by 100 to obtain the percent composition of nitrogen in the sample.
2.112 g N6.807 g sample×100=31.026%
