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what is the percent composition of CuBr2

 Mar 13, 2015

Best Answer 

 #1
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Cu= 63.546

Br=79.904

2Br=2(79.904)=159.808

Wt of molecule= $${\mathtt{63.546}}{\mathtt{\,\small\textbf+\,}}{\mathtt{159.808}} = {\mathtt{223.354}}$$

%Cu=$${\frac{{\mathtt{63.546}}}{{\mathtt{223.354}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{28.450\: \!800\: \!075\: \!216\: \!920\: \!2}}$$

%Br= $${\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{28.45}} = {\frac{{\mathtt{1\,431}}}{{\mathtt{20}}}} = {\mathtt{71.55}}$$

or

$${\frac{{\mathtt{159.808}}}{{\mathtt{223.354}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{71.549\: \!199\: \!924\: \!783\: \!079\: \!8}}$$

 Mar 13, 2015
 #1
avatar
+5
Best Answer

Cu= 63.546

Br=79.904

2Br=2(79.904)=159.808

Wt of molecule= $${\mathtt{63.546}}{\mathtt{\,\small\textbf+\,}}{\mathtt{159.808}} = {\mathtt{223.354}}$$

%Cu=$${\frac{{\mathtt{63.546}}}{{\mathtt{223.354}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{28.450\: \!800\: \!075\: \!216\: \!920\: \!2}}$$

%Br= $${\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{28.45}} = {\frac{{\mathtt{1\,431}}}{{\mathtt{20}}}} = {\mathtt{71.55}}$$

or

$${\frac{{\mathtt{159.808}}}{{\mathtt{223.354}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{71.549\: \!199\: \!924\: \!783\: \!079\: \!8}}$$

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