Two problems of Chemistry

Here are two questions I need help with

Duralin is an alloy of aluminum containing
4.5 percent manganese 3.5 percent copper,
and 0.45 percent magnesium. How much
magnesium is present in a 71 g sample of
this alloy?
Answer in units of g

Duralin is an alloy of aluminum containing
4.5 percent of manganese, 3.5 percent of copper, and 0.45 percent magnesium. How much
copper is present in a 44 g sample of this
alloy?
Answer in units of atom

just use the periodic table to find the amount of grams per mol of each element

100 percent of duralin will have

100 percent - 4.5 - 3.5 - 0.45 = 91.45 percent alluminum

i just subtracted the other percentages to find the whole

we know the mass 71 g

Al = 0.9145*71 = 64.93 g
Mg = 0.0045 * 71 = 0.3195 g
Mn = 0.045 * 71 = 3.195
Cu = 0.035* 71 = 2.485


if you add these up you get the amounts

this is how you find the answer for part 1

Cu = 63.545 g/mol

44 g of duralin has

Cu = 44*0.045 = 1.98

using significant figures we round to

since we know 1 mol equals 63.545 g

x/1 = 1.98/63.545

x = 0.03116

this is the number of mols of Cu

avogadros number says since one mol has 6.023*10^23 atoms

n = 0.03116* 6.023*10^23 = 1.877* 10^22 atoms per 44 g of duralin

hope that helps

71*0.045 = answer 1

44*0.035 * 6.02 E23/(AMU of Cu = 63.5) = answer 2