A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empirical formula of the compound

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H:  0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.