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

ch3 [ch2]14cooh

explanation:

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates