Question

Use dhe References to access inportaat valbes ifneeded for ais A sample of 2.601 grams of a compound containing carbon and hydrogen reacts with oxygen at elevated temperatures to yield 8.793 grams of CO2 and 1.800 grams of H2O. (a) Calculate the masses of C and H in the sample Grams: C- H- (b) Does the compound contain any other elements? (e) What are the mass percentages of C and H in the compound? Mass percentages: C % ; H-| (d) What is the empirical formula of the compound? Esler the elements in the order presested in the question empirical formula-

Answer 1

Weight of compound = 2.601 g

Weight of CO₂ obtained = 8.793 g

Since, 1 mol of CO₂ will be given by 1 mol of C atom present in the compound.

=> 44 g of CO₂ will be given by 12 g of C

=> 8.793 g of CO₂ will be given by (12 ÷ 44) × 8.793 = 2.398 g.

Now, weight of H₂O = 1.8 g

1 mol of H₂O will be given by 2 mol of H atom present in the compound.

=> 18 g of H₂O will be given by 2.016 g of H

=> 1.8 g of H₂O will be given by (2.016 ÷ 18) × 1.8 = 0.202 g

Total weight of C and H = 2.398 + 0.202 g = 2.600 g.

a) Therefore, amount of C present = 2.398 g

Amount of H present = 0.202 g

b) This compound does not contain any other element. It is clear from the amount of C and H that we have calculated.

c) mass % of C = (amount of carbon in compound ÷ total mass of the compound) × 100

Mass % of C = (2.398 ÷ 2.601) × 100 = 92.2%

Mass % of H = (amount of H present in compound ÷ total mass of compound) × 100

Mass % of H = (0.202 ÷ 2.601) × 100 = 7.8%

d) Calculation of empirical formula:

Ans Element Element of age Atomic Mass Atom Mass 98.2 12 1 7- 7.7 7.8 Ratio 707= 77 7-8 = 7.7 1. Н 7.8 . Empirical formula = (CH)