Question

How many grams of phosphorus are in a sample of Ca3(PO4)2 that contains 50.0 grams of calcium

Answer 1

Molar mass of calcium=40
numbers of mole of calcium=50/40/3=5/12
number of mole of phosphorus=(5/12)*2=5/6
mass of phosphoros=(6/5)*31=25.83g//

Answer 2

Answer : The mass of phosphorous present in sample is 25.8 grams.

Explanation :

First we have to calculate the moles of $Ca$.

Molar mass of $Ca$ = 40 g/mole

$\text{ Moles of }Ca=\frac{\text{ Mass of }Ca}{\text{ Molar mass of }Ca}=\frac{50.0g}{40g/mole}=1.25moles$

Now we have to calculate the moles of phosphorous.

In the molecule $Ca_3(PO_4)_2$, there are 3 moles of calcium, 2 moles of phosphorous and 8 moles of oxygen.

As, 3 mole of calcium contains 2 moles of phosphorous

So, 1.25 moles of calcium contains $\frac{1.25}{3}\times 2=0.833$ moles of phosphorous

Now we have to calculate the mass of phosphorous.

$\text{ Mass of phosphorous}=\text{ Moles of phosphorous}\times \text{ Molar mass of phosphorous}$

Molar mass of phosphorous = 31 g/mole

$\text{ Mass of phosphorous}=(0.833moles)\times (31g/mole)=25.8g$

Therefore, the mass of phosphorous present in sample is 25.8 grams.