Question

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of Zn (in at%), and (c) of Pb (in at%)

Answer 1

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

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