1. Hi there! I'm going over some past papers for a university and found this question...unfortunately they don't give the ANSWERS

The problem is: Apatite (Ca5(PO4)3F) is used to manufacture fertilizer. How many kilograms of pure phosphorus (P) is included in 1150 kg of apatite?

I found the molar masses to be as follows:
Ca: 40.08
P: 30.974
O: 15.999
F: 18.998

What I did was added them up as shown in the formula (i.e. 40.08 x 5 + (30.974 + 15.999 x 4) x 3 + 18.998 = 504.308) then I divided the amount of phosphorous (3 x 30.974) by the total amount (504.308) and got approx. 0.184 ...I'm not sure if that step was right but I kept going! Since the amount of phosphorous makes up 0.184 of the total amount I went 1150(kg) x 0.184 and got approx. 211.9 kilograms.

SO basically my answer is 211.9 kg but I have no idea if it is right or completely wrong! Can someone please help me?

Thank you all!

2.

3. Originally Posted by aic18
Hi there! I'm going over some past papers for a university and found this question...unfortunately they don't give the ANSWERS

The problem is: Apatite (Ca5(PO4)3F) is used to manufacture fertilizer. How many kilograms of pure phosphorus (P) is included in 1150 kg of apatite?

I found the molar masses to be as follows:
Ca: 40.08
P: 30.974
O: 15.999
F: 18.998

What I did was added them up as shown in the formula (i.e. 40.08 x 5 + (30.974 + 15.999 x 4) x 3 + 18.998 = 504.308) then I divided the amount of phosphorous (3 x 30.974) by the total amount (504.308) and got approx. 0.184 ...I'm not sure if that step was right but I kept going! Since the amount of phosphorous makes up 0.184 of the total amount I went 1150(kg) x 0.184 and got approx. 211.9 kilograms.

SO basically my answer is 211.9 kg but I have no idea if it is right or completely wrong! Can someone please help me?

Thank you all!
This must be a pretty feeble university. Sounds more like homework. Check in your textbook and see what your teacher has to say.

4. I don't have a professor, these are just practice questions -.-