Calculating Percent Mn in Sample with Mn2(SO4)3 and Al2(SO4)3

[higherme]

A 5.62-g sample containing only Mn2(SO4)3(s) and Al2(SO4)3(s) is dissolved in aqueous solution. The metal ions are precipitated as hydroxides, which are pyrolyzed in air to produce 2.06 g of a mixture of Mn2O3(s) and Al2O3(s).

What was the percent Mn in the original sample?

what i did so far:

found molar mass of Mn2O3 and Al2O3
157.873g/mol and 101.961 g/mol respectively
total 259.834g/mol

Mn2O3 takes up 60.76% of the total molar mass and Al2O3 takes up 39.24%.

so does that mean 60.76% of the 2.06 g mixture would be Mn2O3 and 39.24% would be from Al2O3?

Then i don't know what to do next

help?

 

[higherme]

A 5.62-g sample containing only Mn2(SO4)3(s) and Al2(SO4)3(s) is dissolved in aqueous solution. The metal ions are precipitated as hydroxides, which are pyrolyzed in air to produce 2.06 g of a mixture of Mn2O3(s) and Al2O3(s).

What was the percent Mn in the original sample?

what i did so far:

found molar mass of Mn2O3 and Al2O3
157.873g/mol and 101.961 g/mol respectively
total 259.834g/mol

Mn2O3 takes up 60.76% of the total molar mass and Al2O3 takes up 39.24%.

so does that mean 60.76% of the 2.06 g mixture would be Mn2O3 and 39.24% would be from Al2O3?

Then i don't know what to do next

 

[Gokul43201]

A 5.62-g sample containing only Mn2(SO4)3(s) and Al2(SO4)3(s) is dissolved in aqueous solution. The metal ions are precipitated as hydroxides, which are pyrolyzed in air to produce 2.06 g of a mixture of Mn2O3(s) and Al2O3(s).

What was the percent Mn in the original sample?

what i did so far:

found molar mass of Mn2O3 and Al2O3
157.873g/mol and 101.961 g/mol respectively
total 259.834g/mol

"total molar mass" is a meaningless quantity, unless you are told that the two compounds are in equimolar proportion. They are not.

Assume there are x moles of Mn2(SO4)3 in the original sample. Write down the number of moles of Al2(SO4)3 in terms of x. From here, you can then write down the masses of Mn2O3 and Al2O3 in terms of x.

 

[chemisttree]

A 5.62-g sample containing only Mn2(SO4)3(s) and Al2(SO4)3(s) is dissolved in aqueous solution. The metal ions are precipitated as hydroxides, which are pyrolyzed in air to produce 2.06 g of a mixture of Mn2O3(s) and Al2O3(s).

What was the percent Mn in the original sample?

what i did so far:

found molar mass of Mn2O3 and Al2O3
157.873g/mol and 101.961 g/mol respectively

Yes.

total 259.834g/mol

Mn2O3 takes up 60.76% of the total molar mass and Al2O3 takes up 39.24%.

No, no and no. Total molar mass has absolutely no meaning. You cannot add formula weights like this and get anything useful. Using this logic, binary mixtures of these two species can only occur in one ratio! Therefore, all that follows cannot be correct...

so does that mean 60.76% of the 2.06 g mixture would be Mn2O3 and 39.24% would be from Al2O3?

Then i don't know what to do next

help?

Use the information you are given to find the number of moles of Mn and Al in the original sample. Use those mole #'s to determine the mass of each component using the formula weights of the original sulfates.

 

[higherme]

how to find the moles if i don't know how much from the 2.06g is from Mn2O3 and how much from Al2O3

 

[higherme]

how do i get the moles of Al2(SO4)3 assuming x is moles of Mn2(SO4)3?

 

[higherme]

are they 1:1 ratio?

 

[chemisttree]

Let x = # grams of Mn2O3 and y = # grams of Al2O3
then x + y = 2.06g
It is also true that y = # grams of Al2O3 = 2.06g - x

The number of moles of Mn is 2*(grams Mn2O3)/157.873 and the number of moles of Al2O3 is 2*(grams Al2O3)/101.961 or 2*(2.06g-x)/101.961. These two equations are now expressed in terms of only one unknown, 'x'.

Can you see where this is leading?

 

[higherme]

so...
i'm finding the mass of Mn2O3... and then after that i will know the mass of Mn.. then i can find out of the original sample the % Mn

but to figure out the mass of Mn2O3... do i assume that the moles of Mn and Al are the same?

2*(2.06g-x)/101.961 = 2*(grams Mn2O3)/157.873

 

[higherme]

2*(2.06g-x)/101.961 = 2*(x)/157.873

solve for x?

 

[chemisttree]

so...
i'm finding the mass of Mn2O3... and then after that i will know the mass of Mn.. then i can find out of the original sample the % Mn

but to figure out the mass of Mn2O3... do i assume that the moles of Mn and Al are the same?

No. (why is that message too short?)

 

[chemisttree]

What is the formula weight for Mn2(SO4)3? For Al2(SO4)3?

 

[higherme]

X i found = 1.251639046 = grams of Mn2O3

then find mass of Mn
1.2516g * (1mol Mn2O3/157.873g) * (2mol Mn/1mol Mn2O3) * (54.938g Mn/1mol)
= 0.87111 g Mn

then i see the % that is from the original sample of 5.62 g

(0.87111g / 5.62g) *100%
15.5 %

argh... according to the computer... the answer is not correct...

 

[higherme]

molar mass of Mn2(SO4)3 is 398.014 g/mol
For Al2(SO4)3 is 342.102 g/mol

 

[higherme]

i'm so confused

 

[chemisttree]

From your analysis of the oxides, you know that

moles of Mn is 2*(grams Mn2O3)/157.873 and that (1)
moles of Al is 2*(grams Al2O3)/101.961 or 2*(2.06g-x)/101.961. (2)

Further, you know that the ratio of the number of moles of Mn to Mn2(SO4)3 is 1:2 and likewise Al to Al2(SO4)3 is 1:2. You also know that

the number of grams of Al2(SO4)3 plus number of grams Mn2(SO4)3 is 5.62 grams. (3)

If you were to express the number of grams of Al2(SO4)3 as [(moles Al)*2(moles Al/mol Al2(SO4)3] / [formula wt. of Al(SO4)3]

and

express the number of grams of Mn2(SO4)3 as [(moles Mn)*2(moles Mn/mol Mn2(SO4)3] / [formula wt. of Mn2(SO4)3],

you would be able to solve equation (3) in terms of 'x'. Solve for 'x'.