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

To find the empirical formula of dimethylhydrazine, we calculate the mass of carbon, hydrogen, and nitrogen from the combustion products. We then convert these masses to moles and obtain their simplest whole number ratio to derive the empirical formula.

Step-by-step explanation:

To determine the empirical formula of dimethylhydrazine, we will analyze the combustion products and use them to find the amount of carbon, hydrogen, and nitrogen in the original compound.

Step 1: Determine the amount of Carbon

From the given 0.458 g of CO2, we can calculate the mass of carbon in the CO2 using the molar mass of CO2 (44.01 g/mol). The molar mass of carbon is 12.01 g/mol.

Step 2: Determine the amount of Hydrogen

The 0.374 g of H2O contains a certain mass of hydrogen, which can be calculated by using the molar mass of H2O (18.015 g/mol) and the fact that there are 2 moles of hydrogen atoms for every mole of H2O.

Step 3: Determine the amount of Nitrogen

A 0.486 g sample of dimethylhydrazine produced 0.226 g of N2. Using the molar mass of N2 (28.02 g/mol), we can find the mass of nitrogen in the sample.

Step 4: Calculate the Empirical Formula

We then convert the masses of carbon, hydrogen, and nitrogen to moles and find their simplest whole number ratio to get the empirical formula of dimethylhydrazine.

Answer 2

CH₄N

Step-by-step 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 =
`(mass of CO_2)/(molarmass)*(1 mole of C)/(1 mole of CO_2)`

=
`(0.458)/(44)*(1mole of C)/(1 mole of CO_2)`

= 0.0104 mole

numbers of moles of hydrogen from H2O =
`(mass of H_2O)/(molarmass)*(2 mole of H)/(1 mole of H_2O)`

=
`(0.374)/(18.02)*(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 =
`(0.226)/(0.486)*1`

in 0.312 g of the sample, the nitrogen content will be;
`(0.226)/(0.486)*0.312`

= 0.1450 g of N2

number of moles of N2 =
`(mass)/(molar mass)* (2 mole)/(1 mole)`

=
`(0.1450)/(28.0134) *(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
`(0.0104)/(0.0103)`
`(0.0415)/(0.0103)`
`(0.0103)/(0.0103)`

of moles

1 : 4 : 1

∴ The empirical formula = CH₄N