# Question #afaf3

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The trick here is to realize that *by mass* is **equivalent** to *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 *by number of atoms*, you can say that out of **every**

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

#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.*