Question #3015d

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Question #3015d

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Answer 1 · 2015-11-11T01:03:47

$C u_{2} O$ $Cu_2O$

Explanation:

First, we need to find the amount in moles of each substance. WE have the mass for each so we use the formula:

$n = \frac{m}{M}$ $n = m/M$

Where $n$ $n$ is the amount of the substance in moles, $m$ $m$ is the mass of the substance, and $M$ $M$ is the molar mass or molecular mass of the substance, which we can find by looking at the atomic mass on a periodic table so:

$n (Cu) = \frac{8.882 g}{63.546 g m o l^{- 1}} = 0.139772763 m o l$ $n("Cu") = (8.882g)/(63.546gmol^-1) = 0.139772763mol$

and:

$n (O) = \frac{10 g - 8.882 g}{15.9994 g m o l^{- 1}} = 0.06987762041 m o l$ $n(O) = (10g-8.882g)/(15.9994gmol^-1) = 0.06987762041mol$

Now we find the ratio of the copper and oxygen by dividing both values by the lowest value:

$C u : O = \frac{0.139772763 m o l}{0.06987762041 m o l} : \frac{0.06987762041 m o l}{0.06987762041 m o l}$ $Cu:O = (0.139772763mol)/(0.06987762041mol):(0.06987762041mol)/(0.06987762041mol)$

$= 1.999535219 : 1$ $= 1.999535219 : 1$

The value of 1.999535219 can be rounded to 4 significant figures to get a ratio of $2 : 1$ $2:1$ , as the figures in the question are given to 4 s.f.

From this ratio we can get the empirical formula of $C u_{2} O$ $Cu_2O$

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Question #3015d

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