Questions: 1. How many moles of tetraphosphorus decaoxide are present in 3.21 grams of this compound? moles 2. How many grams of tetraphosphorus decaoxide are present in 4.03 moles of this compound? grams

Solution Steps

Tetraphosphorus decaoxide has the chemical formula \( \text{P}_4\text{O}_{10} \). To find its molar mass, we need to calculate the sum of the atomic masses of all the atoms in the formula:

• Phosphorus (P): \( 4 \times 30.9738 \, \text{g/mol} = 123.8952 \, \text{g/mol} \)
• Oxygen (O): \( 10 \times 15.999 \, \text{g/mol} = 159.99 \, \text{g/mol} \)

Adding these together gives the molar mass of \( \text{P}_4\text{O}_{10} \):

\[ 123.8952 \, \text{g/mol} + 159.99 \, \text{g/mol} = 283.8852 \, \text{g/mol} \]

To find the number of moles of tetraphosphorus decaoxide in 3.21 grams, use the formula:

\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the known values:

\[ \text{moles} = \frac{3.21 \, \text{g}}{283.8852 \, \text{g/mol}} \approx 0.0113 \, \text{moles} \]

To find the mass in grams of 4.03 moles of tetraphosphorus decaoxide, use the formula:

\[ \text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)} \]

Substituting the known values:

\[ \text{mass (g)} = 4.03 \, \text{moles} \times 283.8852 \, \text{g/mol} \approx 1143.0594 \, \text{g} \]

Final Answer