Question

How many grams of `"Pt"` are present in 25.0 g of Cisplatin, an anti-tumor agent?

`Pt(NH_3)_2Cl_2`

Molar mass: `" 300. g mol"^(-1)`

Answer 1

`"16.3 g Pt"`

Explanation:

The idea here is that you need to use the molar mass of cisplatin, `"Pt"("NH"_3)"Cl"_2`, to determine the compound's percent composition of platinium.

As you can see by looking at the compound's chemical formula, one mole of cisplatin contains

• one mole of platinum, `1 xx "Pt"`
• two moles of nitrogen, `2 xx "N"`
• six moles of hydrogen, `6 xx "H"`
• two moles of chlorine, `2 xx "Cl"`

You also know that cisplatin has a molar mass of `"300. g mol"^(-1)`, which means that one mole of cisplatin has a mass of `"300 g"`.

Platinum has a molar mass of `"195.08 g mol"^(-1)`, which means that one mole of platinum has a mass of `"195.08 g"`.

Since `"300. g"` of cisplatin, i.e .one mole of cisplatin, contain `"195.08 g"` of platinum, i.e. one mole of platinum, you can say that `"100 g"` of cisplatin will contain

`100color(red)(cancel(color(black)("g cisplatin"))) * "195.08 g Pt"/(300. color(red)(cancel(color(black)("g cisplatin")))) = "65.03 g Pt"`

Since percent composition tells you the mass of a given element per `"100 g"` of compound, you can say that cisplatin is `65.03%` platinum.

This means that your `"25.0-g"` sample will contain

`25.0color(red)(cancel(color(black)("g cisplatin"))) * overbrace("65.03 g Pt"/(100color(red)(cancel(color(black)("g cisplatin")))))^(color(purple)("65.03% Pt")) = color(green)(|bar(ul(color(white)(a/a)"16.3 g Pt"color(white)(a/a)|)))`

The answer is rounded to three sig figs.

Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin

`25.0 color(red)(cancel(color(black)("g cisplatin"))) * "195.08 g Pt"/(300. color(red)(cancel(color(black)("g cisplatin")))) = color(green)(|bar(ul(color(white)(a/a)"16.3 g Pt"color(white)(a/a)|)))`