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Tatiana [17]
2 years ago
6

What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma

ss? Insert subscripts as needed. empirical formula: HCO HCO
Chemistry
1 answer:
Montano1993 [528]2 years ago
4 0

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

                     1 g of H ----------------- 1 mol

                   3,25 g of H -------------  x

                     x = (3.25 x 1) / 1

                     x = 3.25 moles

                    12 g of C ---------------- 1 mol

                     19.36 g of C ----------  x

                     x = (19.36 x 1) / 12

                     x = 1.61 moles

                     16g of O --------------- 1 mol

                     77.39 g of O ---------  x

                      x = (77.39 x 1)/16

                      x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

                        C₂HO₃

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A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol
Lunna [17]
(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M


8 0
2 years ago
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