The compound is CH6N2, has molar mass of 46 g/mol
26.1 % carbon
13.2 % hydrogen
60.7 % Nitrogen
Explanation:
Complete combustion of 35,0 mg of the compound pro-
duced 33,5 mg CO, and 41.1 mg H,O. What is the empirical formula of the compound?
Since the compound contains only C, H and N. This compound has the formule CxHyNz
Calculate moles of CO2
Moles CO2 = Mass Co2 / Molar mass CO2
Moles CO2 = 0.0335 grams / 44.01 g/mol
Moles CO2 = 0.00076 moles = 0.76 mmoles
Every mole CO2 contains 1 mole of carbon and 2 moles of oxygen
This means 0.76 mmoles of CO2 contains 0.76 mmoles of carbon
Calculate mass of carbon in the compound:
Mass carbon = moles carbon * Molar mass carbon
Mass carbon = 0.00076 moles * 12.01 g/mol
Mass carbon in the compound = 0.00913 grams =9.13 mg
Calculate % carbon in the compound
Total mass of the compound = 35.0 mg
% carbon = (9.13 mg/35.0 mg) *100%
% carbon = 26.1%
Calculate moles H2O
35 mg of the combound is combusted to produce 41.1 mg of H2O
Moles H2O = mass H2O / Molar mass H2O
Moles H2O = 0.0411 grams / 18.02 g/mol
Moles H2O = 0.00228 moles = 2.28 mmoles
Every mole H2O contains 2 moles hydrogen and 1 mole oxygen
This means 2.28 mmoles H2O contains 2*2.28 = 4.56 mmoles of hydrogen
Calculate mass of hydrogen in the compound
Mass hydrogen = Moles hydrogen * Molar mass hydrogen
Mass hydrogen = 0.00456 moles * 1.01 g/mol
Mass of hydrogen = 0.00461 grams = 4.61 mg
Calculate % hydrogen in the compound
% hydrogen = (4.61 / 35.0)*100 = 13.2 %
Calculate % N in the compound
100% - 26.1% - 13.2 % = 60.7 %
60.7 % = 21.245 mg
Number of moles N = 0.021245 / 14 g/mol = 0.00152 moles
The compound has
0.00076 moles carbon = 26.1 %
0.00456 moles hydrogen = 13.2 %
0.00152 moles Nitrogen = 60.7 %
To find the empirical formule. We divide each number of mole by the smallest amount of mole. (This is Carbon, 0.00076)
Carbon: 0.00076/ 0.00076 = 1
Hydrogen: 0.00456/0.00076 = 6
Nitrogen: 0.00152 / 0.00076 = 2
This means the empirical formula of the compound is CH6N2
ii. A 65.2-mg sample of the compound was analyzed for
nitrogen by the Dumas method, giving 35.6 mL of dry N, at 740, torr and 25°C.
The partial pressure of N2 = 740 torr ( this is 740/760 torr = 0.974 atm
The volume of N2 = 35.6 mL = 0.0356 L
The temperature = 25°C = 273.15 + 25 = 298.15 Kelvin
We can find the number of moles via the ideal gas law p*V=n*R*T
⇒ with p = the pressure of the gas in atm = 0.974 atm
⇒ with V= the volume of the gas in L = 0.0356 L
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L*atm/ K*mol
⇒ with T = the temperature in Kelvin = 298.15 Kelvin
n = (p*V)/(R*T)
n = (0.974 * 0.0356)/(0.08206*298.15)
n = 0.00142 moles = 1.42 mmoles
Calculate mass of nitrogen in the compound:
0.00142 moles *28.02 g/mol = 0.03978 grams = 39.78 mg
% Nitrogen = (39.78 / 65.2)*100 % = 60.7 %
The effusion rate of the compound as a gas was mea
sured and found to be 24.6 mL/min. The effusion rate of argon gas, under identical conditions, is 26.4 mL/min. What is the molecular formula of the compound?
The graham's law of effusion syas:
(Rate of effusion for gas 1)/(Rate of effusion for gase 2) = √M2/√M1
with M1 = Molar mass of gas 1 ; M2 = molar mass of gas 2
24.6/26.4 = √39.948/√M1
M1 = 46 g/mol
This confirms the answers from the previous questions. The compound is CH6N2, has molar mass of 46 g/mol
26.1 % carbon
13.2 % hydrogen
60.7 % Nitrogen
