Acombustion analysis of 5.214 g of a compound yields 5.34 g co 2 , 1.09 g h 2 o, and 1.70 g n 2 . if the molar mass of the compound is 129.1 g/mol, what is the chemical formula of the compound?

Let the formula of compound be . The mass and molar mass of compound is 5.214 g and 129.1 g/mol respectively.

The combustion of compound gives 5.34 g of , 1.09 g of and 1.70 g of . First number of moles of carbon, hydrogen, oxygen and nitrogen to compare the molar ratio.

Calculation for number of moles:

For number of moles of C, first calculate number of mole of :

Here, m is mass and M is molar mass.

Molar mass of is 44.01 g/mol thus,

Since, 1 mol of have 1 mole of C thus, number of C will be 0.1213 mol.

Or,

Convert this into mass as follows:

Molar mass of C is 12 g/mol thus,

For number of moles of H, first calculate the number of moles of :

Molar mass of is 18 g/mol thus,

Since, 1 mol of have 2 mole of H thus, number of H will be 2×0.060 mol=0.12 mol.

Or,

Convert this into mass as follows:

Molar mass of H is 1 g/mol thus,

For number of moles of N, first calculate the number of moles of :

Molar mass of is 28 g/mol thus,

Since, 1 mol of have 2 mole of N thus, number of N will be 2×0.060 mol=0.12 mol.

Or,

Convert this into mass as follows:

Molar mass of N is 14 g/mol thus,

Since, mass of compound is 5.214 g thus, mass of oxygen will be:

Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:

The ratio of number of moles of C, H,O and N will be:

Thus, empirical formula of compound will be CHON.

According to above formula molar mass of compound will be 43 g/mol

Now, according to chemical formula, the molar mass is 129.1 g/mol taking the ratio:

Thus, chemical formula will be

Therefore, chemical formula of compound is .

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

c (diamond),

sio2 (quartz)