Question

I have mass of the precipitate, moles of MgNH4PO4*6H2O (the precipitate) and I need to find Moles of P. Help!

Answer 1

@Abmon98

Answer 2

multiply the mass percent composition of P by the moles of the compound.

\(\sf mass ~percent ~composition~ of ~P=\dfrac{Molar~mass ~of~ P}{Molar~mass~of~compound}\)

Answer 3

what's the mass percent composition of P? and by compound do you mean MgNH4PO4*6H2O? @aaronq

Answer 4

the mass percent composition is the ratio of the molar masses of P to the whole compound. (yes by compound i mean MgNH4PO4*6H2O). You have to calculate the number (See formula above).

Answer 5

ok I think I've got those. It also wants the Mass of P2O5 in the plant food. I'm attaching my chart to this @aaronq

Answer 6

you can get their molar masses from the periodic table

Answer 7

How would that help? I'm a little confused sorry

Answer 8

sorry, i read what you wrote wrong.

So you need the mass of P2O5? Did you do anything else with the precipitate, like react it with something else?

Answer 9

no the precipitate just formed after the lab and I weighed it. @aaronq

Answer 10

They should be cristals right? like star cristals?

Answer 11

non-color right?

Answer 12

it looks like white hardened clay almost. It's very small.

Answer 13

you got it from MgCl2 in ammonial solution?

Answer 14

plant food, ammonia, epsom salt, and water

Answer 15

Oh so you mixed those things and collected the precipitate Mg..?

Answer 16

yep!

Answer 17

look, first wee need to get only the P, once you get that you would get PO5 by combustion of P

Answer 18

Yeah, we need the moles of P. Did you find those earlier?

Answer 19

@dsood15 ?

Answer 20

When you get them, divide them by 2 because of the stoichiometry.

\(P_2O5 \rightarrow 2P..\)

Then multiply the moles by the molar mass of \(P_2O_5\) to get the mass.

moles*molar mass =mass

Answer 21

Pretty much everything I have is in that chart I attached. @aaronq

Answer 22

ok so to that you need to add a clorhidric solution aprox. 5 N... that's gonna get you all the PO4 in the precipitate as we dont have ay VO3 nor WO4

Answer 23

huh

Answer 24

ammm... also add a .05 mol solution of ZrOCl2... the white precipitate should be Zr3(PO4)4

Answer 25

So say, for the first trial, you have moles of P =0.00089

To find the moles of P2O5 (because of the stoichiometry) we divide by 2.

Now we multiply the moles by the molar mass

0.00089 moles *283.89 g/mol = 0.2526621 grams \(\approx\)0.253 g

Answer 26

but you said to divide by two so is it
0.00089/ 2 * 283.89 ?

Answer 27

.126ish?

Answer 28

@aaronq

Answer 29

oh, yep. sorry about that

Answer 30

ow... I thought you needed the actual P and P2O5

Answer 31

How do I find the percentage @aaronq

Answer 32

I know i was confused for a while. dsood, you gotta give a little more background on your questions. were not mind readers.

The percentage =\(\sf \dfrac{mass~of~P_2O_5}{mass~of~sample}*100\%\)

Answer 33

haha sorry about that didn't mean to... ok I'll try that and see how it goes. thanks for all your help @aaronq

Answer 34

it should work out. no problem, dude !