Stöchiometrie Prüfungsaufgabe?

26.36 g of carbon dioxide contained:

m(O) = 26.36 g * 32/44 = 19.171 g

m(C) = 26.36 g * 12/44 = 7,1891 g

The 44 is the numerical value of the molar mass of CO2, 12 of C and 32 of 2 O. The units are getting out.

This calculates the mass of hydrogen in the unknown substance.

m(H) = 8.4 g – 7.1891 g = 1.21 g

For the ratio and sum formula, the ratio of the quantities n is required. This is calculated with n = m/M

n(C) = 7,1891 g/(12 g/mol) = 0.59909 mol

n(H)=1.21 g/l (g/mol)=1.21 mol

Now it is necessary to expand skillfully so that both values provide as close as possible a ratio of integers. Here you can see that the quantity of H is quite exactly the double quantity of C. The empirical formula:

CH₂ or in general C_nH_2n

The molar mass of the hydrocarbon is added to 84.16 g/mol. This fits

C₆H₁₂

It could therefore be, inter alia, a witch or cyclohexane. However, if the molar mass of the unknown hydrocarbon is already known anyway, nothing has to be reckoned with, since only one sum formula is possible.