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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: peroxide62 on January 13, 2017, 12:53:13 AM

Title: Binary Mixture
Post by: peroxide62 on January 13, 2017, 12:53:13 AM

Amounts of tin may be alloyed with copper to produce bronzes of varying composition and properties.
4.4464 g of a particular bronze, compounded of only Cu and Sn, can be quantitatively converted into a 5.5860-g mixture of the oxides CuO and SnO2.

What was the percentage (by mass) of tin in the alloy?
 %

I did it but it's wrong. What did I did wrong? And how do I do this question?

Z g= Cu   Y g= Sn

Zg x (1mol Cu/ 63.53gCu) x (79.35g CuO/ 1mol CuO) = 1.2525X g Cu

Yg x (1mol Sn/ 118.69g Sn) x ( 150.69g SnO2/1 mol of SnO2) = 1.2696Y g Sn


1.2525X + 1.2696Y = 5.586
          X + Y = 4.4464.           

1.2525(4.4464-Y) + 1.2696Y = 5.586
5.569 - 1.2525Y + 1.2696Y = 5.586
0.0171Y = 0.0166
Y=0.9707

0.9707/4.4464 x 100% = 21.83%

Title: Re: Binary Mixture
Post by: AWK on January 13, 2017, 01:43:04 AM

Quote

Zg x (1mol Cu/ 63.53gCu) x (79.35g CuO/ 1mol CuO) = 1.2525X g CuO

Yg x (1mol Sn/ 118.69g Sn) x ( 150.69g SnO2/1 mol of SnO2) = 1.2696Y g SnO₂

Quote

0.0171Y = 0.0166

Use correct masses and do correct subtraction.