Question

multi - step molar conversions
show your work. include units round final answers to 3 sig figs.
keep 2 digits after the decimal place on all molar masses.
give answers in scientific notation where appropriate.

1. there are ( 3.20\times10^{22} ) atoms of copper in the outer shell of pennies. how many grams of

copper is this?

1. 124 grams of calcium oxide contains how many formula units?
2. a tiny crystal of table salt (nacl) contains 950. billion formula units of salt. what is the mass

of this crystal, in grams?

1. what is the mass (in grams) of ( 8.40\times10^{21} ) molecules of water?

Explanation:

Problem 1

Step1: Determine molar mass of copper

The chemical formula of copper is \(Cu\). The molar mass of \(Cu\) is \(63.55\ g/mol\).

Step2: Use Avogadro's number (\(N_A = 6.022\times10^{23}\ atoms/mol\))

First, find the number of moles of \(Cu\). The number of moles \(n=\frac{3.20\times 10^{22}\ atoms}{6.022\times 10^{23}\ atoms/mol}\)

$$n=\frac{3.20}{6.022}\times10^{- 1}\ mol\approx0.0531\ mol$$

Then, find the mass \(m\) using \(m = n\times M\) (where \(M\) is molar mass). So \(m=0.0531\ mol\times63.55\ g/mol\)

$$m = 0.0531\times63.55\ g\approx3.38\ g$$

Step1: Determine molar mass of calcium oxide (\(CaO\))

The molar mass of \(CaO\): \(M=(40.08 + 16.00)\ g/mol=56.08\ g/mol\)

Step2: Find the number of moles of \(CaO\)

\(n=\frac{124\ g}{56.08\ g/mol}\approx2.21\ mol\)

Step3: Use Avogadro's number (\(N_A = 6.022\times10^{23}\ formula\ units/mol\))

The number of formula units \(N=n\times N_A\)
\(N = 2.21\ mol\times6.022\times 10^{23}\ formula\ units/mol\)

$$N=2.21\times6.022\times10^{23}\approx1.33\times 10^{24}\ formula\ units$$

Step1: Convert 950. billion to scientific notation

\(950. billion=950\times10^{9}=9.50\times 10^{11}\ formula\ units\)

Step2: Determine molar mass of \(NaCl\)

The molar mass of \(NaCl\): \(M=(22.99 + 35.45)\ g/mol = 58.44\ g/mol\)

Step3: Find the number of moles of \(NaCl\)

\(n=\frac{9.50\times 10^{11}\ formula\ units}{6.022\times 10^{23}\ formula\ units/mol}\)

$$n=\frac{9.50}{6.022}\times10^{-12}\ mol\approx1.58\times10^{-12}\ mol$$

Step4: Find the mass \(m\) using \(m=n\times M\)

\(m = 1.58\times10^{-12}\ mol\times58.44\ g/mol\)

$$m=1.58\times58.44\times10^{-12}\ g\approx9.24\times10^{-11}\ g$$

Answer:

The final answer (rounded to 3 sf with units) is \(3.38\ g\)

Problem 2