Question

Question #be6a5

Answer 1

`1.2 * 10^(24)"e"^(-)`

Explanation:

Your strategy here will be to

• use calcium's molar mass to calculate how many moles of calcium are present in your sample `color(red)((1))`
• use Avogadro's number to convert the moles to number of atoms `color(blue)((2))`
• use the atomic number of calcium to figure out how many electrons you have in the sample `color(darkgreen)((3))`

`color(red)((1))` Grams of calcium to moles of calcium

Calcium has a molar mass of `"40.08 g mol"^(-1)`, which means that one mole of calcium has a mass of `"40.08 g"`. Use this as a conversion factor to help you find the number of moles present in your `"4 g"` sample

`4 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.08color(red)(cancel(color(black)("g")))) = "0.0998 moles Ca"`

`color(blue)((2))` Moles of calcium to atoms of calcium

As you know, in order to have one mole of an element, you need to have `6.022 * 10^(23)` atoms of that element `->` this is known as Avogadro's number and is essentially the definition of a mole.

`color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"atoms"color(white)(a/a)|)))`

You can use Avogadro's number as a conversion factor to figure out how many atoms of calcium are present in your sample. You will have

`0.0998 color(blue)(cancel(color(black)("moles Ca"))) * (6.022 * 10^(23)"atoms Ca")/(1color(blue)(cancel(color(black)("mole Ca")))) = 6.01 * 10^(22)"atoms Ca"`

`color(darkgreen)((3))` Atoms of calcium to number of electrons

In order to determine how many electrons you have in the sample, you need to use the number of electrons present in one neutral atom of calcium.

Calcium has an atomic number equal to `20`, which means that a neutral calcium atom has `20` protons inside its nucleus and `20` electrons surrounding its nucleus.

Use this as a conversion factor to find the number of electrons present in your sample

`6.01 * 10^(22) color(darkgreen)(cancel(color(black)("atoms Ca"))) * "20 e"^(-)/(1color(darkgreen)(cancel(color(black)("atom Ca")))) = color(darkgreen)(|bar(ul(color(white)(a/a)color(black)(1.2 * 10^(24)"e"^(-))color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the mass of calcium.