What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

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eyelesshamster eyelesshamster · Answer 1 · 2017-12-14T20:00:01+01:00

What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

Na2CO3

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kobenhavn kobenhavn · Answer 2 · 2019-07-20T17:21:29+02:00

Answer: [tex]Na_2CO_3[/tex].

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 = 43.38 g

Mass of C = 11.33 g

Mass of O = 45.29 g

Step 1 : convert given masses into moles.

Moles of Na=[tex]\frac{\text{ given mass of Na}}{\text{ molar mass of Na}} \frac{43.38g}{23g/mole}=1.89moles[/tex]

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{11.33g}{12g/mole}=0.94moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{45.29g}{16g/mole}=2.83moles[/tex]

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

For Na = [tex]\frac{1.89}{0.94}=2[/tex]

For C =[tex]\frac{0.94}{0.94}=1[/tex]

For O =[tex]\frac{2.83}{0.94}=3[/tex]

The ratio of Na : C: O = 2: 1: 3

Hence the empirical formula is [tex]Na_2CO_3[/tex].