# Question #07c92

##### 1 Answer

#### Answer:

#### Explanation:

So, you know that you can get copper metal by roasting a mixture that contains *cuprite*, *copper(II) sulfide*,

The keys to this problem will be the balanced chemical equations for the reactions that take place.

Both compounds will react with oxygen gas to form copper metal and sulfur dioxide

#"Cu"_2"S"_text((s]) + "O"_text(2(g]) -> color(red)(2)"Cu"_text((s]) + "SO"_text(2(g])#

and

#"CuS"_text((s]) + "O"_text(2(g]) -> "Cu"_text((s]) + "SO"_text(2(g])#

Now, it's worth noting that the question goes out of its way to try and distract you by using *purity*.

For example, instead of telling that you are heating

Likewise, you're interested in how much **pure** copper is produced by the two reactions, so use the given purity to determine how much copper you have

#75.2color(red)(cancel(color(black)("g impure Cu"))) * "89.4 g Cu"/(100color(red)(cancel(color(black)("g impure Cu")))) = "67.23 g Cu"#

Now, let's assume that the initial sample contains

#x + y = "89.0 g" " " " "color(purple)((1))#

The *molar masses* of cuprite, copper(II) sulfide, and copper metal, respectively, are

#"Cu"_2"S " -> " 159.16 g/mol"# #"CuS " -> " 95.611 g/mol"# #"Cu " -> " 63.546 g"#

If you have

#xcolor(white)(x)color(red)(cancel(color(black)("g"))) * ("1 mole Cu"_2"S")/(159.16color(red)(cancel(color(black)("g")))) = x/159.16"moles Cu"_2"S"#

Likewise, if you have

#ycolor(white)(x)color(red)(cancel(color(black)("g"))) * ("1 mole CuS")/(95.611color(red)(cancel(color(black)("g")))) = y/63.546"moles CuS"#

The number of moles of copper will be

#67.23color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "1.058 moles Cu"#

Now focus on the mole ratios that exist between the two species in the mixture and the copper metal.

Notice that you have a

#x/159.16color(red)(cancel(color(black)("moles Cu"_2"S"))) * (color(red)(2)" moles Cu")/(1color(red)(cancel(color(black)("mole Cu"_2"S")))) = (2x)/159.16"moles Cu"#

Do the same for the

#x/95.611color(red)(cancel(color(black)("moles CuS"))) * ("1 mole Cu")/(1color(red)(cancel(color(black)("mole CuS")))) = x/95.611"moles Cu"#

You know that the moles of copper formed by the first reaction and the moles of copper formed by the second reaction **must add up** to give

This means that you have

#(2x)/159.16 + y/95.611 = 1.058" " " "color(purple)((2))#

From this point on, you have a classic system of two equations with two unknowns. Use equation

#x = 89.0- y#

then plug this into equation

#(2(89.0-y))/159.16 + y/95.611 = 1.058#

This will be equivalent to

#17019 - 191.22y + 159.16y = 16100.1#

#32.06y = 918.9 implies y = 918.9/32.06 = "28.7 g"#

The value of

#x = 89.0 - 28.7 = "60.3 g"#

Therefore, the percent composition of the mixture will be - remember to take the **total mass of the mixture**, including *impurities*!

#(28.7color(red)(cancel(color(black)("g"))))/(100.0color(red)(cancel(color(black)("g")))) xx 100 = color(green)("28.7%")# #-># copper(II) sulfide

and

#(60.3color(red)(cancel(color(black)("g"))))/(100.0color(red)(cancel(color(black)("g")))) xx 100 = color(green)("60.3%")# #-># cuprite