Percentage Composition Question (1 Viewer)

[renton]

A student analysed a 2.85g sample of washing powder for its phosphorous content.

The phosphorous was precipitated as Mg2P207 and then filtered.

The mass of the precipitate was 0.125g

Determine the percentage by mass, of phosphorous in the washing powder.

Taken from some 2002 trial paper, think its independant schools

 

[rama_v]

Find number of moles of precipitate. Then find number of moles of phosphorus using the molar ratio (from the formula we see that there are two moles of phosphorous atoms for every mole of ppt). Then you can find the grams of phoshorous and hence % comp

 

[tristambrown]

"A student analysed a 2.85g sample of washing powder for its phosphorous content.

The phosphorous was precipitated as Mg2P207 and then filtered.

The mass of the precipitate was 0.125g

Determine the percentage by mass, of phosphorous in the washing powder.

Taken from some 2002 trial paper, think its independant schools"

If the student had precipitated pure P then: percentage P would be 0.125/2.85 = 0.04385964912 = (100/1 for perentage) 4.39%

Unfortunatly we arent that lucky as the precip is a compound with several different elements in it.

we have to calculate the molar weight of the precipitated compound
work out how many moles of the compound we have
work out how many moles of P are in that specific compound
multiply our molar weight by the weight of P to work out our mass of P
then we have to divide our mass of p by the original sample mass to find how much of original sample was P


Mg = 24.305 G /mol .. 2 are called for accoding to formula = 48.61 g/mol
P = 30.974 g/mol .. 2 are called for = 61.948 g/mol
o = 15.999 g/mol ..7 are called for =111.993

total weight for 1 mol of Mg2P207 =111.993 + 61.948 + 48.61 =225.551g

weight of sample is 0.125g
therefore moles = samle weight / molar weight
.125/225.551 = 0.00056166901 moles of Mg2P207 but within this there are 2 moles of Mg, 2 Moles of P and 7 Moles of O so to find molar amt P we need to multiply Mg2P207 by 2

0.00056166901 * 2 = moles of Phosphorous = 0.00112333802

Weight (mass) of one mole P = 30.974

Moles we have (0.00056166901) * weight for 1 mole (30.974)= mass P we have = 0.03479427187g P

Our mass of P/ mass of original washing powder sample will give us Pther ratio of initial sample
0.03479427187g (ie 0.03479427187(P):99.98778803(other stuff)

= 0.1220851645

* 100/1 for percentage
=1.220851645
= 1.22%


This looks very small so im kind of worried ive made a mistake, but the original maximum value of 4% also seems to agree with this low percentage so either ive buggered both calculations up, or there is really not very much P in this sample at all

whats the real answer ?

 

[renton]

i'd love to tell you, but i don't have the answers. sorry bout that, but thanks for your help anyway.