Part A A compound contains 0.483 g N and 1.26 F. The molar mass of the compound

Question

Part A

A compound contains 0.483 g N and 1.26 F. The molar mass of the compound is 104 g/ mol. what are the molecular and empirical formulas of the compound?

Part B What is the concentration of an Ba (OH)2(aq) solution which requires 25.00 ml to titrate 12.30 ml of 0.0250 M H3PO4 (aq)?

3 Ba(OH)2(aq)+ 2 H3PO4 (aq)->Ba3(PO4)2(s)+6 H2O (l)

Part C What Mass of Hydrogen gas will be produced from 4.76 g of NO and 3.23 g of CH4? 2 NO (g) + 2 CH4 (g)-> 2 HCN (g) 2 H2O (g) +H2 (g)

Part D

A 250.0 Sample of a Metal with specific heat 0.95 J/g*°C is heated to 80.0 °C and then placed in a calorimeter with 325.0 g of water. The initial temperature of the water is 19.0 °C. The heat loss to the calorimeter is negligible. What is the final temperature of the water?

Explanation / Answer

Part A: Since the compound contains 0.483 g N and 1.26 g F, % of N = 0.583 X 100 /(0.483 + 1.26) = 27.71.

% of F = 1.26 X 100 /(0.483 + 1.26) = 72.29.

Relative number of atoms of N = 27.71/mass no. = 27.71/14 = 2

Relative number of atoms of F = 72.29/mass no. = 72.29/19 = 4

Ratio of number of N atoms = 2/2 =1

Ratio of number of F atoms = 4/2 = 2

Therefore the emperical formula is NF2

Emperical formula mass = 14 + 2 X 19 = 14 + 38 = 52

Molecular mass = 104

n = molecar formula mass / emperical formula mass = 104/52 = 2

Molecular formula = (NF2)2 = N2F4

Part B: Here we make use of the normality eqn. N1 X V1 = N2 X V2

N1 = normality of Ba(OH)2

V1 = Volume of Ba(OH)2 = 25 mL

N2 = normality of H3PO4 = Molarity X Basicity = 0.0250 X 3 = 0.0750N

V2 = volume of H3PO4 = 12.30 mL

N1 = N2 X V2/N1 = 0.075 X 12.30/25 = 0.0369N

Molarity of Ba(OH)2 = N1 X acidity = 0.0369 / 2 = 0.0185M

Part C: Molecular mass of NO = 14 + 16 = 30

Molecular mass of CH4 = 12 + 4 = 16

Molecular mass of H2 = 2

According to the eqn. 2 X 30g of NO combines with 2 X 16g of CH4 to give 2g of H2.

or

1g of NO combines with 32g/60 of CH4 to give 2g/60 of H2

or

4.76g of NO combines with 2.54g of CH4 to give 0.16g of H2

3.23-2.54 = 0.69g of CH4 will remain unreacted.

Part D: Data insufficient. Unit of the weight of metal sample not mentioned.

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