An organic compound containing only c, h, and n yields the
following data
i. complete combustion of 35,0 mg of the compound pro-
duced 33,5 mg co, and 41.1 mg h, o.
ii. a 65.2-mg sample of the compound was analyzed for
nitrogen by the dumas method (see exercise 129),
giving 35.6 ml of dry n, at 740, torr and 25°c.
ili. 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 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

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

contour lines on a map indicate elevation, or depth. numeric notations on contour lines refer to the level of elevation, or height about sea level.