(0.334g/2.000g)*100= ???
4-significant figures.
The active ingredient in some antiperspirants is aluminum chlorohydrate, Al2(OH)5Cl. Analysis of a 2.000-g sample of antiperspirant yields 0.334 g of aluminum. What percent (by mass) of aluminum chlorohydrate is present in the antiperspirant? (Assume that there are no other compounds containing aluminum in the antiperspirant.)
2 answers
No. Anonymous has given the % Al in the compound.
g Al2(OH)5Cl = 0.334g Al x (molar mass Al2(OH)5Cl/atomic mass Al) = ?
Then % Al2(OH)5Cl = [g Al2(OH)5Cl/2.000)]*100 = ? and it is to 3 s.f. based on the 0.334 g Al.
g Al2(OH)5Cl = 0.334g Al x (molar mass Al2(OH)5Cl/atomic mass Al) = ?
Then % Al2(OH)5Cl = [g Al2(OH)5Cl/2.000)]*100 = ? and it is to 3 s.f. based on the 0.334 g Al.
3 answers