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What is the empirical formula for a compound that contains 88.8 percent copper and 11.2 percent oxygen?
Wiki User
∙ 11y ago
Empirical formula is Cu:O::88.8%: 11.2%
Divide each percentage by the respective atomic masses.
Hence
88.8/64.5 : 11.2/16 = 1.376 : 0.7
Then divide by the lowest ratio
1.376 / 0,7 : 0.7 /0.7 =1.965 : 1 ~ 2:`1
Hence
Cu:O :: 2:1
Hence empericiasl formula is Cu2O (NB THis is copper in oxidatiob state '1').
lenpollock
Wiki User
∙ 15y agoHaven't got a calculator on me, but I'll get you started. 88.8% O = .888 g O 11.2% H = .112 g H .888 g O * (1 mol O / 15.999 g O) = aboutish .0555 mol O .112 g H * (1 mol H / 1.0079 g H) = about .112 mol H That's the ratio of moles of O to H, divide both of them by the smaller of the two, .0555, and it looks like you'll wind up with almost whole numbers, so just round them and those are the coefficients for that element.
Wiki User
∙ 10y agoC3H5 or C5H3
Wiki User
∙ 12y agoAtomic weights of hydrogen and oxygen
Wiki User
∙ 11y agoCo2O Copper (I) Oxide
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