a.f.x.d
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In an experiment, 3.425 g of lead oxide was reduced to form 3.105 g of lead.

The empirical formula of the lead oxide is

A PbO
B Pb3O2
C Pb3O4
D Pb4O3


The answer is C but I don't know-how. can someone explain this?
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EierVonSatan
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I like your creative thread titles :lol:

What is the mass of oxygen that is lost during decomposition?
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a.f.x.d
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(Original post by EierVonSatan)
I like your creative thread titles :lol:

What is the mass of oxygen that is lost during decomposition?
it's not given in the question so I suppose we have to subtract the mass of lead from lead oxide and get 0.32g
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EierVonSatan
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(Original post by a.f.x.d)
it's not given in the question so I suppose we have to subtract the mass of lead from lead oxide and get 0.32g
Correct

So now you have the mass of oxygen: 0.32g
And the mass of lead: 3.105g

Can you now find the empirical formula from there?
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a.f.x.d
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(Original post by EierVonSatan)
Correct

So now you have the mass of oxygen: 0.32g
And the mass of lead: 3.105g

Can you now find the empirical formula from there?
I'm getting A as the answer whereas it's C in the marking scheme. could you please explain?
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EierVonSatan
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(Original post by a.f.x.d)
I'm getting A as the answer whereas it's C in the marking scheme. could you please explain?
mol of oxygen: mass/Mr = 0.32 / 16 = 0.02
mol of lead: mass/Mr = 3.105/207 = 0.015

Now, divide by the smallest (0.015) to get: Pb = 1 and O = 4/3

Multiply both by 3 to remove the fraction: Pb3O4

(I assume you've rounded 4/3 down to 1, which might be okay at GCSE but not A-level)
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scimus63
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try looking here, its gcse but will probably help

https://www.science-revision.co.uk/m...20formula.html
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a.f.x.d
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(Original post by EierVonSatan)
mol of oxygen: mass/Mr = 0.32 / 16 = 0.02
mol of lead: mass/Mr = 3.105/207 = 0.015

Now, divide by the smallest (0.015) to get: Pb = 1 and O = 4/3

Multiply both by 3 to remove the fraction: Pb3O4

(I assume you've rounded 4/3 down to 1, which might be okay at GCSE but not A-level)
thank you so much!
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