percentage in compound when compound is not known?

Question

The combustion method used to analyze for carbon and hydrogen can be adapted to give percentage N by collecting the nitrogen from combustion of the compound as N2. A sample of a compound weighing 8.66 mg gave 1.72 mL N2 at 24C and 748 mmHg. What is the percentage N in the compound?

Answer 1

1) It seems that we need to use the Gas Ideal Law:

PV = nRT ......(Eq.1)

2) Using the given data:

V=1.72 mL = 0.00172 L
T = 24°C = 297.15 K
P= 748 mmHg
with the units above, the constant gas would be:
R = 62.36 L mmHg mol^-1 K^-1

we can know the number of moles of N2 substituting in the equation above (Eq.1):

(748 mmHg)(0.00172 L) = n (62.36 L mmHg mol^-1 K^-1 )(297.15K)

solving for "n":

n = (748 mmHg)(0.00172 L) / 62.36 L mmHg mol^-1 K^-1 )(297.15K)

n= 6.943 x 10^-5 moles of N2

3) Then, we can know the mass of Nitrogen from the found number of moles above (remember that molecular mass M of N2 is 28 g/mol):

m = nM = (6.943 x 10^-5 moles N2)(28 g/mol N2) =1.944x10^-3 grames of N2 = 1.94 mg of N2

4)As we know the mass of the sample m = 8.66 mg, we can say that the percentage of N2 in it is:

% N2 = (1.94 mg/8.66 mg)x100% = 22.4%

That´s it!

Good luck!

Answer 2

Firstly, you need to convert from mass to molar abundance. Taking the atomic weights of each atom: C = 12; H = 1; O = 16 Divide the mass percentage by the numbers above. Eg: 41.3% C, 55.2% O, 3.5% H 41.3% / 12 = 3.44 55.2% / 16 = 3.45 3.5% / 1 = 3.50 Now, you need to see what most closely resembles an integer mix of numbers. In this case, I've chosen 1:1:1, but you'd have to play around. Bear in mind that there's big differences between different compounds with the same proportions: both graphite and diamond are pure 100% Carbon...

Answer 3

H form H2O
C CO2
N NO2

O = 100% - % H2O - CO2% -NO2%