A compound is found to consist of 34.5% sodium, 16.4% boron, and 48.6% oxygen. What is the empirical formula?

Chemistry

1 answer:

NARA [144]2 years ago

7 0

Answer: The empirical formula is $NaBO_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given:

Mass of Na= 34.5 g

Mass of B= 16.4 g

Mass of O = 48.6 g

Step 1 : convert given masses into moles.

Moles of Na = $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{34.5g}{23g/mole}=1.5moles$

Moles of B = $\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{16.4g}{11g/mole}=1.5moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.6g}{16g/mole}=3moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Na = $\frac{1.5}{1.5}=1$

For B = $\frac{1.5}{1.5}=1$

For O = $\frac{3}{1.5}=2$

The ratio of Na: B: O= 1: 1: 2

Hence the empirical formula is $NaBO_2$

You might be interested in

The value of Ka for phenol (a weak acid), C6H5OH, is 1.00×10-10. Write the equation for the reaction that goes with this equilib

I am Lyosha [343]

Answer:

$Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Explanation:

Hello,

In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$

Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:

$K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$