Question

Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a 0.312 gg sample yields 0.458 gg CO2CO2 and 0.374 gg H2OH2O. The nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.What is the empirical formula of dimethylhydrazine?Express your answer as a chemical formula.

Answer 1

Answer:

CH₄N

Explanation:

Given that;

mass of the sample =  0.312 g

mass of CO2 = 0.458 g

mass of H2O =  0.374 g

nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.

Let start with calculating the respective numbers of moles of Carbon Hydrogen and Nitrogen from the given data.

numbers of moles of Carbon from CO2 = $\frac{mass of CO_2}{molarmass}*\frac{1 mole of C}{1 mole of CO_2}$

= $\frac{0.458}{44}*\frac{1mole of C}{1 mole of CO_2}$

= 0.0104 mole

numbers of moles of hydrogen from H2O = $\frac{mass of H_2O}{molarmass}*\frac{2 mole of H}{1 mole of H_2O}$

= $\frac{0.374}{18.02}*\frac{2 mole of H}{1 mole of H_2O}$

= 0.02077 × 2

= 0.0415 mole

The nitrogen content of a 0.486 g sample is converted to 0.226 g N2

Now, in 1 g of the sample; The nitrogen content = $\frac{0.226}{0.486}*1$

in 0.312 g of the sample, the nitrogen content will be; $\frac{0.226}{0.486}*0.312$

= 0.1450 g of N2

number of moles of N2 = $\frac{mass}{molar mass}* \frac{2 mole}{1 mole}$

= $\frac{0.1450}{28.0134} *\frac{2 mole}{1 mole}$

= 0.0103 mole

Finally to determine the empirical formula of Carbon  Hydrogen and Nitrogen; we have:

                                Carbon         Hydrogen           Nitrogen

number of moles      0.0104         0.0415                 0.0103

divided by the

smallest number      $\frac{0.0104}{0.0103}$             $\frac{0.0415}{0.0103}$                    $\frac{0.0103}{0.0103}$

of moles

                                   1        :           4           :               1

∴ The empirical formula = CH₄N