Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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]

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.