Question

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene. Upon complete combustion in oxygen, you collect 0.204 g of CO2 and 0.030 g H2O (we’ll assume that these are the products, along with nitrogen). TNT is made up solely of C, H, O, and N, and it contains 18.5% nitrogen by mass. The empirical formula of TNT is;

Answer 1

The empirical formula is C7H5N3O6

Step-by-step explanation:

Step 1: Data given

Mass of sample = 0.150 grams

Mass of CO2 = 0.204 grams

Molar mass CO2 = 44.01 g/mol

Mass of H2O = 0.030 grams

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.204 grams / 44.01 g/mol

Moles CO2 = 0.00464 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.00464 moles we have 0.00464 moles C

Step 4: Calculate mass C

Mass C = 0.00464 moles * 12.01 g/mol

Mass C = 0.0557 grams

Step 5: Calculate moles H2O

Moles H2O = 0.030 grams / 18.02 g/mol

Moles H2O = 0.00166 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00166 moles H2O we have 2* 0.00166 = 0.00332 moles H

Step 7: Calculate mass H

Mass H = 0.00332 moles * 1.01 g/mol

Mass H = 0.00335 grams

Step 8: Calculate mass N

Mass N = 0.185 * 0.150 grams

Mass N = 0.02775 grams

Step 9: Calculate moles N

Moles N = 0.02775 grams / 14.0 g/mol

Moles N = 0.00198 moles

Step 10: Calculate mass O

Mass O = 0.150 grams - 0.02775 - 0.00335 - 0.0557

Mass O = 0.0632 grams

Step 11: Calculate moles O

Moles O = 0.0632 grams / 16.0 g/mol

Moles O = 0.00395 moles

Step 11: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.00464 moles / 0.00198 moles =2.33

H: 0.00332 moles / 0.00198 moles = 1.66

N: 0.00198 moles / 0.00198 moles = 1

O: 0.00395 moles / 0.00198 moles = 2

For 1 mol N we have 2.33 moles C, 1.66 moles H and 2 moles O

OR

For 3 moles N we have 7 moles C, 5 moles H and 6 moles O

The empirical formula is C7H5N3O6