Calculating Empirical Formula for Dimethylhydrazine Rocket Fuel

[Larrytsai]

Dimethylhydrazine, the fuel used in rocket propulsion systems, is made of carbon, hydrogen, and nitrogen atoms. A 2.859 g sample of this compound is burned in excess air, and produces 4.190 g of carbon dioxide and 3.428 g of water. What is the empirical formula? (molar masses: Co2= 44.01g/mol; H2O= 18.02g/mol)

I was wondering do I predict the reactants? and " the compound is burned" what is the compound? I am not use to this question because I am usually given the percents in the compound so i don't know how to start off this question...

K well igot a ratio of 2/1 for H2O and Co2 but i was wondering should i find the percent composition of C and H then add and subtract from 100? to get the percent of N and start the work from there? and to find the percent composition should i find it from CO2(H2O) or seperately?

 

[symbolipoint]

You are looking for calculations about a compound which contains apparently only Carbon, Nitrogen, and Hydrogen. We don't know yet the ratios of each element per forumula unit of compound.

The resulting measured carbon dioxide gives you information about the Carbon quantitiy in hydrazine. The resulting measured water gives you information about the amount of Hydrogen in hydrazine.

That information is an approach to make. The rest is to write a reaction with some variables, and apply stoichiometry and basic algebra.

 

[Kushal]

CxHyNz + O2 -------> xCO2 + (y/2)H2O + a nitrogen cmpd

so 1 mol of CxHyNz gives x mol of CO2.
n CO2 = 0.0952 mol

x = 0.0952

1 mol CxHyNz also gives (y/2) mol of H2O
n H2O = 0.1902 mol

y/2 = 0.1902 mol
y = 0.3804 mol

you have CxHyNz weighing 2.859g

you already have Ar C, H and N and x and y. you can find z

(12*0.0952) + (1*0.3804) + (14*z) = 2.859

with x, y and z you can calculate the empirical formula.