Solveeit Logo

Question

Question: A \(2\cdot 38\)g of uranium is strongly heated in a current of air. The resulting oxide will weigh \...

A 2382\cdot 38g of uranium is strongly heated in a current of air. The resulting oxide will weigh 28062\cdot 806g. Determine the empirical formula of compound.

Explanation

Solution

First find the mass of oxygen present in the resulting oxide by subtracting the given mass of uranium from the oxide. Then find the no. of moles for uranium and oxygen by dividing their given masses in oxide by their atomic masses. Obtain the simple whole no. ratio to get the empirical formula.

Complete step by step answer:
The given mass of uranium in the oxide is 2382\cdot 38g and the mass of oxide is 28062\cdot 806g. So, the mass of oxygen in this resulting oxide = mass of oxide – mass of uranium
\begin {align} &\text{Mass of oxygen in oxide} =2\cdot 806-2\cdot 38 \\\ & =0\cdot 426 \\\ \end{aligned}
Finding the no. of moles for uranium we have, no. of moles = given mass/molar mass
No. of moles of uranium in oxide =238238=001=\dfrac{2\cdot 38}{238}=0\cdot 01 (atomic mass of uranium =238=238)
No. of moles of oxygen in oxide =042616=0027=\dfrac{0\cdot 426}{16}=0\cdot 027 (atomic mass of oxygen is=16=16)
Converting these ratios into simple ratios by dividing both the ratios with the smallest ratio between the two i.e. with 0010\cdot 01
So, for uranium we get 001001=1\dfrac{0\cdot 01}{0\cdot 01}=1
And for oxygen we get 0027001=27\dfrac{0\cdot 027}{0\cdot 01}=2\cdot 7
For empirical formula both the ratios should be the whole number ratio so multiplying 272\cdot 7 with 33 to get its nearest whole number ratio and correspondingly the ratio of uranium should also be multiplied by 33
Therefore we get, simple whole no. ratio for uranium =1×3=3=1\times 3=3
And simple whole no. ratio for oxygen=27×3=8=2\cdot 7\times 3=8
So, the formula for resulting oxide includes 33moles of uranium and 88moles of oxygen i.e. U3O8{{U} _{3}} {{O}_ {8}} .

Additional Information:
Empirical formula tells us about the simplest ratio of elements in the compound whereas molecular formula tells us about the total no. of atoms of an element present in the compound. The empirical formula for two compounds can be similar as it is the simplest ratio of its constituent elements but two different compounds can never have the same molecular formula.

Note:
The empirical formula should always be in the whole number values. The method to find empirical formula includes, finding the no. of moles of constituent elements and then converting them into simple whole number ratios. Molecular formula provides more information about a compound than empirical formula because it tells us about the exact no. of atoms of an element present in the compound. One should study about the atomic masses for different elements.