Question

1)A mixture of butene, C4H8 , and butane,C4H10, is burned in air to give CO2 and water. Suppose you burn 2.84 g of the mixture and obtain 8.77 g of CO2 and 3.99 g of H2O . What are the mass percentages of butene and butane in the mixture?

Mass percentage of butene = %

Mass percentage of butane = %

2)Copper metal can be prepared by roasting copper ore, which can contain cuprite (Cu2S) and copper(II) sulfide.

Suppose an ore sample contains 11.0% impurity in addition to a mixture of CuS and Cu2S. Heating 100.0 g of the mixture produces 75.1 g of copper metal with a purity of 89.2%. What is the weight percent of CuS in the ore? The weight percent of Cu2S?

Weight percent of = %

Weight percent of = %

Answer 1

The balanced reactions will be

C4H8 + 6O2 ----------------- 4CO2 + 4H2O [Balance the reaction first by carbon count by making coefficient of CO2 to 4, then balance the hydrogen by making coefficient of H2O to 4 and then balance the oxygen]

C4H10 + 6.5O2 ----------------- 4CO2 + 5H2O [Balance the reaction first by carbon count by making coefficient of CO2 to 4, then balance the hydrogen by making coefficient of H2O to 5 and then balance the oxygen]

Let the mass of butene be x

Mass of Butane will be (2.84-x)

Molar mass of CO2 = 44.01 gm/mol

Number of moles of CO2 = Mass/Molar mass = 8.77/44.01 = 0.19927 moles

Both C4H8 amd C4H10 will lead to 4 moles of CO2

Hence, number of moles of C4H8 + number of moles of C4H10 = 0.19927/4 = 0.0498175

Molar mass of C4H8 = 56.1 gm/mol

Molar mass of C4H10 = 58.12 gm/mol

Let the moles of C4H8 be a

Let the moles of C4H10 be b

a + b = 0.0498175

56.1a + 58.12b = 2.84

a= 0.0274226 and b= 0.0223949

Mass of Butene = number of moles * molar mass = 0.0274226 * 56.1 = 1.5384g

Mass of Butane = number of moles * molar mass = 0.0223949 * 58.12 = 1.3015g

Mass % of Butene = 1.5384/2.84 * 100 = 54.17%

Mass % of Butane = 1.3015/2.84 * 100 = 45.83%