Question

1. what do we use to convert between grams and moles (g→mol or mol→g)?
2. convert to moles: 45 g of c.
3. convert to grams: 0.33 moles of caso₄.
4. convert 17 grams of co₂ into molecules of co₂.

Explanation:

Step1: Calculate molar mass of C

The molar mass of \( C \) is \( 12.01\ g/mol \).

Step2: Use formula \( n=\frac{m}{M} \)

Given \( m = 45\ g \) and \( M=12.01\ g/mol \), then \( n=\frac{45}{12.01}\approx3.75\ mol \).

Step1: Calculate molar mass of \( CaSO_{4} \)

Molar mass of \( Ca = 40.08\ g/mol \), \( S = 32.07\ g/mol \), \( O = 16.00\ g/mol \).
\( M(CaSO_{4})=40.08 + 32.07+4\times16.00=136.15\ g/mol \)

Step2: Use formula \( m = n\times M \)

Given \( n = 0.33\ mol \), then \( m=0.33\times136.15 = 44.93\ g \approx45\ g \)

Step1: Calculate molar mass of \( CO_{2} \)

Molar mass of \( C = 12.01\ g/mol \), \( O = 16.00\ g/mol \).
\( M(CO_{2})=12.01+2\times16.00 = 44.01\ g/mol \)

Step2: Calculate moles of \( CO_{2} \)

Using \( n=\frac{m}{M} \), with \( m = 17\ g \), \( n=\frac{17}{44.01}\approx0.386\ mol \)

Step3: Calculate number of molecules

Using \( N=n\times N_{A} \), \( N_{A}=6.022\times10^{23}\ mol^{-1} \)
\( N = 0.386\times6.022\times10^{23}\approx2.32\times10^{23} \)

We use the molar mass (\( M \)) of a substance. The formula \( n=\frac{m}{M} \) (for \( g
ightarrow mol \)) and \( m = n\times M \) (for \( mol
ightarrow g \)) where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M \) is the molar mass in \( g/mol \).

Answer:

\( 3.75\ mol \)