Answer :
Answer: [tex]K_2SO_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 K = 49.4 g
Mass of S = 20.3 g
Mass of O = 30.3 g
Step 1 : convert given masses into moles.
Moles of K=[tex]\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{49.4g}{40g/mole}=1.23moles[/tex]
Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]
Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For K = [tex]\frac{1.23}{0.63}=2[/tex]
For S =[tex]\frac{0.63}{0.63}=1[/tex]
For O =[tex]\frac{1.89}{0.63}=3[/tex]
The ratio of K: S:O = 2: 1: 3
Hence the empirical formula is [tex]K_2SO_3[/tex].
The empirical formula of a compound that contains 49.4% K, 20.3% S, and 30.3% by mass is option B, [tex]\bold{K_2SO_3}[/tex].
What is an empirical formula?
The empirical formula of a compound is the total number of molecules of each element in that compound.
Given,
The masses
49.4 g K, 20.3 g S, and 30.3 g.
Step 1: To calculate the moles
[tex] \bold{Number\;of \;moles= \dfrac{mass}{molar\;mass}}[/tex]
Moles of potassium K
[tex]\bold{Number\;of \;moles\;of\;K = \dfrac{49.4\;g}{39.9\;u}= 1.23\;moles}[/tex]
Moles of S
[tex]\bold{Number\;of \;moles\;of\;S = \dfrac{20.3g}{32g/mole}=0.63moles[/tex]
Moles of O
[tex]\bold{Number\;of \;moles\;of\;S = \dfrac{30.3g}{16g/mole}=1.89moles[/tex]
Step 2: To calculate the mole ratio by dividing each mole by the smallest value of mole calculated.
For K
[tex]\bold\dfrac{1.23\;moles}{0.63\;moles}= 1.95\;moles}[/tex]
For S
[tex]\bold\dfrac{0.63moles}{0.63moles}= 1\; mole}[/tex]
For O
[tex]\bold\dfrac{1.89moles}{0.63moles}= 3\; mole}[/tex]
The ratio of K: S:O = 2: 1: 3
Thus, the empirical formula is [tex]\bold{K_2SO_3}[/tex]. Correct option is B.
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