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)#
Molar mass:
1 Answer
Explanation:
The idea here is that you need to use the molar mass of cisplatin,
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
Platinum has a molar mass of
Since
#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
This means that your
#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)|)))#