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Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.
A.
KBH
B.
KB2H4
C.
KB3H9
D.
K2B5H9

Sagot :

MeenaMe
Use the percent as mass in g 
Mass g) --> g/mol 
K, B and H molar mass from periodic table

54.93 g K x (1 mol K/ 39.10 g K) = 1.405 mol K
38.73 g B x (1 mol B/ 10.81 g B) = 3.583 mol B
6.34 g H x (1 mol H/ 1.008 g H) = 6.290 mol H

Divide all three answers by the smallest value and you will get 
1.000 mol K
2.550 mol B
4.477 mol H 
now multiply these three answers by a number that will make all a whole number or (a number with a 9 as the first decimal point)
so multiply by 2
2 mol k
5 mol B
and 9 mol H
E.F. = K2B5H9 
Answer is D 
Sorry for bad explanation!!!!!!


dexteright02

Hello!

Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.

A.     KBH

B.     KB2H4

C.     KB3H9

D.     K2B5H9

  • We have the following data:

Potassium (K) ≈ 39 a.m.u (g/mol)

Boron (B) ≈ 11 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

  • We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

K: 54.93 % = 54.93 g

B: 38.73 % = 38.73 g

H: 6.34 % = 6.34 g

  • The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

[tex]K: \dfrac{54.93\:\diagup\!\!\!\!\!g}{39\:\diagup\!\!\!\!\!g/mol} \approx 1.408\:mol[/tex]

[tex]B: \dfrac{38.73\:\diagup\!\!\!\!\!g}{11\:\diagup\!\!\!\!\!g/mol} \approx 3.521\:mol[/tex]

[tex]H: \dfrac{6.34\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 6.34\:mol[/tex]

  • We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

[tex]K: \dfrac{1.408}{1.408}\to\:\:\boxed{K = 1}[/tex]

[tex]B: \dfrac{3.521}{1.408}\to\:\:\boxed{B \approx 2.5}[/tex]

[tex]H: \dfrac{6.34}{1.408}\to\:\:\boxed{H \approx 4.5}[/tex]

convert number of atomic radio into whole number

2 * (1 : 2.5 : 4.5)

= 2 : 5 : 9  ← whole number of atomic radio

K = 2

B =  5 

H = 9

  • Thus, the minimum or empirical formula found for the compound will be:

[tex]\boxed{\boxed{K_2B_5H_9}}\Longleftarrow(Empirical\:Formula)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]

Answer:

D.     K2B5H9

________________________

[tex]\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]

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