Empiric Formula of Crom Oxid: CrO

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SUMMARY

The empirical formula of chromium oxide (CrO) can be determined from the mass of chromium (1.12 g) and chromium oxide (1.63 g) analyzed. The calculations yield a chromium percentage of 68.7% and an oxygen percentage of 31.3%. The initial ratio of chromium to oxygen is calculated as 1.32:1.96, which simplifies to approximately 1:1.5, leading to the conclusion that the empirical formula is Cr2O3, not CrO. The key step is recognizing the need to further divide the ratios to achieve a simplified whole number ratio.

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danne89
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I can't solve this: 1.63 g crom oxid gave 1.12 g crom during analysis. Calculate the empiric formula of crom oxid.

% Cr = 1.12 : 1.63 * 100 = 68.7 %
% O = 100 - 68.7 = 31.3 %
Cr_xO_y
x:y = 68.7 : Cr : 31.3 : O
= 68.7 : 52 : 31.3 : 16
= 1.32 : 1.96
= 1 : 1
CrO
Which is completely wrong.
 
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Hello, try this: 1.32 and 1.96 can be further divided to give 1 and 1.485; this means that approximately 1:1.5 ratio is present. This is the same as 2 to 3, I mean Cr2O3

To sum up, dividing further to obtain a 1 in the series is the most important step here.
 
Ahh! I understand!
 

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