Question #afaf3

1 Answer
Dec 6, 2015

Answer:

Here's what I got.

Explanation:

The trick here is to realize that #12.3%# tin by mass is equivalent to #12.3%# tin by number of atoms.

You know that the total mass of the sample is determined by the mass of tin and by the mass of lead present in the sample.

In turn, the mass of tin will depend on how many atoms of tine you have in the sample, and the mass of lead will depend on how many atoms of lead you have in the sample.

So, if the sample is #12.3%# tin by number of atoms, you can say that out of every #100# atoms present in the mixture, #12.3# will be tin atoms.

This means that you can use the total number of atoms present in the mixture to find the number of atoms of tin

#7.395 * 10^(22) color(red)(cancel(color(black)("atoms mix"))) * "12.3 atoms Sn"/(100color(red)(cancel(color(black)("atoms mix")))) = 9.096 * 10^(21)"atoms Sn"#

Implicitly, the number of atoms of lead will be

#overbrace(7.395 * 10^(22))^(color(red)("total no. of atoms")) - overbrace(9.096 * 10^(21))^(color(blue)("atoms of Sn")) = 6.485 * 10^(22)"atoms Pb"#

Next, use Avogadro's number to figure out how many moles of tin and how many moles of lead can be found in the mixture.

As you know, one mole of any element will contain exactly #6.022 * 10^(23)# atoms of that element. This means that you have

#9.096 * 10^(21) color(red)(cancel(color(black)("atoms Sn"))) * "1 mole Sn"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Sn")))) = "0.01510 moles Sn"#

and

#6.485 * 10^(22) color(red)(cancel(color(black)("atoms Pb"))) * "1 mole Pb"/(6.022 * 10^(23) color(red)(cancel(color(black)("atoms Pb")))) = "0.1077 moles Pb"#

Finally, to get the masses of the two elements, use their molar masses, which tell you exactly what the mass of one mole of a substance is.

In your case, you will have

#0.01510 color(red)(cancel(color(black)("moles Sn"))) * "118.7 g"/(1color(red)(cancel(color(black)("mole Sn")))) = "1.792 g Sn"#

and

#0.1077 color(red)(cancel(color(black)("moles Pb"))) * "207.2 g"/(1color(red)(cancel(color(black)("mole Pb")))) = "22.32 g Pb"#

The total mass of the sample will thus be

#m_"total" = "1.793 g" + "22.32 g" = color(green)("24.1 g")#

The answer is rounded to three sig figs

SIDE NOTE Interestingly enough, I got a slightly different value for the total mass of the sample. The way I see it, 24.1 g seems like an accurate result, though.