mole calculation of chemistry

1.)What is the mass (in grams) of one molecule of CCl4?

Mass of 1 mole of CCl4 = 12 + 35.5x4 = 154 g
No. of molecules in 1 mole of CCl4 = 6.02 x 1023 molecule
Mass of 1 molecule of CCl4 = 154/(6.02 x 1023) = 2.56 x 10-22 g


2.) How many chlorine atoms are present in 22.1 g of the compound?

No. of moles of CCl4 = 22.1/154 = 0.1435 mol
Each mole of CCl4 contains 4 moles of Cl atoms.
No. of moles of Cl atoms = 0.1435 x 4 = 0.574 mol
Each mole of Cl atoms contains 6.02 x 1023 Cl atoms.
No. of moles of Cl atoms = 0.574 x (6.02 x 1023) = 3.46 x 1023 atoms


3.) A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of H2O.
The molar mass of the compound is 180.156 g/mol.

1.
Molar mass of C = 12 g/mol
Molar mass of CO2 = 12 + 16x2 = 44 g/mol
Mass fraction of C in CO2 = 12/44

Mass of C in 48.64 g of the compound
= Mass of C in CO2 formed
= 71.30 x (12/44) g
= 19.45 g

2.
Molar mass of H = 1 g/mol
Molar mass of H2O = 1x2 + 16 = 18 g/mol
Mass fraction of H in H2O = (1x2)/18 = 2/18

Mass of H in 48.64 g of the compound
= Mass of H in H2O formed
= 29.19 x (2/18)
= 3.24 g

3.
Mass of O in 48.64 g of the compound
= 48.64 - (19.45 + 3.24)
= 25.95 g

[Mole ratio C : H : O = 19.45/12 : 3.24/1 : 25.95/16 = 1 : 2 : 1]
[Empirical formula of the compound = CH2O]
[Let (CH2O)n be the molecular formula of the compound.]
[n(12 + 2x1 + 16) = 180.156]
[n = 6]
[Molecular formula = C6H12O6]