How do you find the empirical formulas for 65.2% Sc, 34.8% O?

Question

How do you find the empirical formulas for 65.2% Sc, 34.8% O?

1 Answer

Impact of this question

6724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T13:29:15

As with all these problems, we ASSUME a $100 \cdot g$ $100*g$ mass of compound.........and come up with an empirical formula of $S c_{2} O_{3}$ $Sc_2O_3$ .

Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species. We assume a $100 \cdot g$ $100*g$ mass of $S c_{2} O_{3}$ $Sc_2O_3$ , and we come up with a molar ratio:

$Moles of scandium$ $"Moles of scandium"$ $=$ $=$ $\frac{Mass of scandium}{Molar mass of scandium}$ $"Mass of scandium"/"Molar mass of scandium"$ $=$ $=$ $\frac{65.2 \cdot g}{44.96 \cdot g \cdot m o l^{- 1}} = 1.45 \cdot m o l$ $(65.2*g)/(44.96*g*mol^-1)=1.45*mol$ .

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{Mass of oxygen}{Molar mass of scandium}$ $"Mass of oxygen"/"Molar mass of scandium"$ $=$ $=$ $\frac{34.8 \cdot g}{16.00 \cdot g \cdot m o l^{- 1}} = 2.175 \cdot m o l$ $(34.8*g)/(16.00*g*mol^-1)=2.175*mol$ .

In each instance, I divided thru by the ATOMIC MASS of the element. And now if we divide thru by the smallest molar quantity, (that of the metal), I get a formula of $S c O_{1.5}$ $ScO_(1.5)$ . Because, by specification, the empirical formula is a WHOLE number ratio, we quote its empirical formula as $S c_{2} O_{3}$ $Sc_2O_3$ . Capisce?

Science

Math

Humanities

... and beyond

How do you find the empirical formulas for 65.2% Sc, 34.8% O?

1 Answer

Explanation:

Related questions

Impact of this question