Molar mass ( chemistry)

a)
P1V1 = P2V2
1.05 atm X 1 litre = 1 atm X V2
V2 = 1.05 litre
24 litre at 1 atm is 1 mole. 1.05 litre sample at 1 atm has mass 4.24g
24 litre sample will has mass = (24/1.05) X 4.24 = 96.9 g
molar mass = 96.9 g/mole

b)----------------------------------------- C -------------------- O ----------------------- Cl
mass (g) /100g sample----------12.1------------------16.2---------------------71.7
no. of mole ------------------------(12.1/12)-----------(16.2/16)------------(71.7/35.5)
---------------------------------------- = 1.008 ----------- = 1.0125 ------------ = 2.02
simplest ratio ---------------- (1.008/1.008) --- (1.0125/1.008) ------ (2.02/1.008)
---------------------------------------- = 1 ----------------- = 1.004 -------------- = 2.004

empirical formula : COCl2
empirical formula mass = 12 + 16 + 35.5 X 2 = 99

96.9 / 99 app. = 1

molecular formula = COCl2

Molecular weight of CO2: 12+32=44g
Molecular weight of H2O: 2+16=18g

1.794g CO2: 1.794/44=0.04 mole
0.729g H2O: 0.729/18=0.04 mole

CO2:H2O = 0.04:0.04
= 1:1
No. of mole of C : H = 1:2

No. of mole of C : 0.04
Mass of C : 0.04x12=0.48g

No. of mole of H : 0.04x2=0.08
Mass of H : 0.08x1=0.08g

Mass of O : 1.224-0.48-0.08=0.664g
No. of mole of O : 0.664/16=0.04

Hence, C:H:O=1:2:1

Empirical formula is CH2O