chemistry questions

1a)
A gaseous compound A composition by mass N 21.6％, O 49.2％, F 29.2％
Molar mass of nitrogen is 14.0067 g/mol
Molar mass of oxygen is 15.9994 g/mol
Molar mass of fluorine is 18.9984 g/mol
A gaseous compound A composition by mole is
= N 21.6/14.0067, O 49.2/15.9994, F 29.2/18.9984
= N 1.542, O 3.075, F 1.5369
= N : O : F = 1 : 2 : 1
The empirical formula of A is NO2F
1b)
PV = nRT
density = m' * ( P / RT )
at 298K & 1 atm, R = 0.082 atm dm^3 /K/mol, density of A is 2.65 g dm/3
m' = density * ( RT / P )
= 2.65 * 8.31441 * 298 / 101
= 65.01 g / mol
The molar mass of A is 65.01 g / mol
For A to be (NO2F)n
molar mass = n * (14.0067 + 15.9994 * 2 + 18.9984) = 64.7554
n * 65.0039 = 65.01
n = 1
The molecular formula of A is NO2F

2a)
2.4 g of ethanoic acid vaporised volume 1.5 dm^3, temperature of 450 K, pressure of 69.8 kPa
PV = nRT
density = m' * ( P / RT )
m' = density * ( RT / P )
m' = 2.4 / 1.5 * (8.31441 * 450 / 69.8)
m' = 85.765 g / mol
the apparent molar mass of ethanoic acid under the experimental conditions is 85.765 g / mol
2b)
We know that ethanoic acid is CH3COOH which have a molar mass of 60.1 g/mol
maybe the sample's actual weight is 1.68g or the actual temperature is 300K
m' = density * ( RT / P )
m' = 1.68 / 1.5 * (8.31441 * 450 / 69.8)
m' = 85.765 g / mol
m' = 60.035 g / mol
m' = density * ( RT / P )
m' = 2.4 / 1.5 * (8.31441 * 300 / 69.8)
m' = 85.765 g / mol
m' = 60.035 g / mol
I hope this can help your understanding. :)

PV = nRT