. Dimethyl nitrosamine is a known carcinogen. It can be formed in the intestinal tract when digestive juices react with the nitrite ion is preserved and smoked meats. It is made up of carbon, hydrogen, nitrogen, and oxygen atoms. A 4.319 g sample of dimethyl nitrosamine burned in oxygen yields 5.134 g of CO2 and 3.173 g of H2 O. The compound contains 37,82% by mass of nitrogen. What is the empirical formula of dimethyl nitrosamine?

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06-09-2019
Chemistry

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. Dimethyl nitrosamine is a known carcinogen. It can be formed in the intestinal tract when digestive juices react with the nitrite ion is preserved and smoked meats. It is made up of carbon, hydrogen, nitrogen, and oxygen atoms. A 4.319 g sample of dimethyl nitrosamine burned in oxygen yields 5.134 g of CO2 and 3.173 g of H2 O. The compound contains 37,82% by mass of nitrogen. What is the empirical formula of dimethyl nitrosamine?

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Alleei Alleei · Answer 1 · 2019-09-12T13:08:32+02:00

Answer: The empirical formula for the given compound is [tex]C_2H_6ON_2[/tex]

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[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and nitrogen respectively.

We are given:

Mass of [tex]CO_2=5.134g[/tex]

Mass of [tex]H_2O=3.173g[/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 5.134 g of carbon dioxide, [tex]\frac{12}{44}\times 5.134=1.4g[/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 3.173 g of water, [tex]\frac{2}{18}\times 3.173=0.35g[/tex] of hydrogen will be contained.

For calculating the mass of nitrogen:

As, 100 g of sample contains 37.87 % of nitrogen

So, 4.319 g of sample contains [tex]\frac{37.87}{100}\times 4.319=1.635g[/tex] of nitrogen

For calculating the mass of oxygen:

Mass of oxygen in the compound = (4.319) - (1.4 + 0.35 + 1.635) = 0.934 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.4g}{12g/mole}=0.116moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.35g}{1g/mole}=0.35moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.934g}{16g/mole}=0.058moles[/tex]

Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1.635g}{14g/mole}=0.116moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.058 moles.

For Carbon = [tex]\frac{0.116}{0.058}=2[/tex]

For Hydrogen = [tex]\frac{0.35}{0.058}=6.03\approx 6[/tex]

For Oxygen = [tex]\frac{0.058}{0.058}=1[/tex]

For Oxygen = [tex]\frac{0.116}{0.058}=2[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : N = 2 : 6 : 1 :2

Hence, the empirical formula for the given compound is [tex]C_2H_6O_1N_2=C_2H_6ON_2[/tex]