Please help me with this empirical formula!?

Question

a compound contains only C, H, and N. Combustion of 42.0 mg of the compound produces 40.2 mg CO2 and 49.3 mg H2O. What is the empirical formula of the compound?

so confused....

Answer 1

There will be the same number of moles of C and H and N on both sides of the equation, so all the atoms that were in your starting sample are now in your products. Carbon from your original sample is now present in CO2, and all H is in the H2O. N is currently unknown, but you can calculate the N once you have done the H and O

What you need to do is work out how many moles of C and H are in the products

Hydrogen

Calculate moles of H in H2O

molar mass H2O = 16.00 + (2 x 1.008) = 18.016 g/mol

molar mass H = 1.008 g/mol

moles = mass / molar mass

moles H2O = 0.0493 g / 18.016 g/mol

= 0.002736 moles of H2O

In ever H2O there are 2 moles of H

therefore moles of H in H2O, and thus in your original hydrocarbon = 2 x 0.002736

= 0.005472 moles of H

Now you can work out mass of H

Mass H = moles x molar mass

= 0.005472 g x 1.008 g/mol

= 0.005516 g of H was in original sample = 5.516 mg

Carbon

Calculate moles C in CO2

molar mass CO2 = (16.00 x 2) + 12.01 = 44.01 g/mol

moles CO2 = mass / molar mass

= 0.0402 g / 44.01 g/mol

= 0.0009134 moles of CO2

Every CO2 molecule has 1 Carbon, therefore moles of C in CO2 and thus in the original hydrocarbon

= 0.0009134  moles of C

Now, mass C = moles x molar mass

= 0.0009134 mol x 12.01 g/mol

= 0.0110 g of Carbon was in original sample = 11.0 mg

Nitrogen:

We know the 42.0 mg of the original had 11.0 mg of Carbon and 5.516 mg of H

So N = total mass - mass C - mass H

= 42.0 - 11.0 - 5.516

= 25.484 mg of N in the original sample

moles of N in original = mass/ molar mass

= 0.025484 g / 14.01 g/mol

= 0.001819 moles of N

Now work out the ratio of C : H : N

moles of C : H : N

0.0009134  :  0.005472 : 0.001819

To get it into whole number divide all numbers in the ratio by the lowest number

C : H : N

(0.0009134/0.0009134)  :  (0.005472/0.0009134) : (0.001819/0.0009134)

ratio = 1 : 6 : 2

Therefore empirical formula = CH6N2