A major motor company displays a die-cast model of its first automobile, made from 9.35kg of iron. To celebrate its one-hundreth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model?
~my thoughts:
m= 9.35kg
atomic mass of gold= 196.97g/mol
atomic mass of iron= 55.845g/mol
9.35kg(1000g/1kg)(1molFe/55.845g)(196.97g/1mol Au)(1kg/1000g)= 33.0kg of Au
~The answer in the book is 23.0kg so I don't know what I did wrong, however I think that the problem is btwn the moles of Fe and moles of Au conversion b/c usually in chem you have a rxn where you say the ratio of the 2 compounds to each other.
I'm not sure where I went wrong
can someone help me out here?
~Thanks~
3 answers
You need to multiply the 9.35 kg by the density ratio, NOT the atomic mass ratio. Densities are not proportional to atomic mass, although they do tend to rise with higher amu numbers. The amu is a measure of the number of protons and neutrons in the nucleus, but for different elements, the interatomic distances also change
Thank you drwls, I got the same answer as the book =D
ρ iron = 7870 kg/m3 m iron =9.35kg
ρ gold =19320 kg/m3
v=m/ρ ρ=m/v
v=9.35/7870
v=1.1880×10^(-3)
m=v×ρ m=1.1880×10^(-3)×19320
m=22.9 kg≈23kg
ρ gold =19320 kg/m3
v=m/ρ ρ=m/v
v=9.35/7870
v=1.1880×10^(-3)
m=v×ρ m=1.1880×10^(-3)×19320
m=22.9 kg≈23kg
1 answer