What is the percent by mass of titanium in ru | vedclass

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(C) In the rutile structure,oxide ions form a closest-packed array. If there are $N$ oxide ions,the number of octahedral holes is $N$. Since titanium

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$ \simeq 60\%, +4 $

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(C) In the rutile structure,oxide ions form a closest-packed array. If there are $N$ oxide ions,the number of octahedral holes is $N$. Since titanium occupies one-half of the octahedral holes,the number of $Ti$ atoms is $N/2$. The ratio of $Ti:O$ is $(N/2):N = 1:2$. Thus,the formula of the mineral is $TiO_2$. The oxidation number of $Ti$ in $TiO_2$ is calculated as $x + 2(-2) = 0$,so $x = +4$. The molar mass of $TiO_2 = 48 + 2(16) = 80 	ext{ g/mol}$. The percentage by mass of $Ti = (48 / 80) 	imes 100 = 60\% $.

What is the percent by mass of titanium in rutile,a mineral that contains titanium and oxygen,if the structure can be described as a closest-packed array of oxide ions,with titanium in one half of the octahedral holes? What is the oxidation number of titanium? $(Ti = 48, O = 16)$

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