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Mariulka [41]
3 years ago
6

A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?

Chemistry
2 answers:
Artemon [7]3 years ago
7 0

Answer:

H₂O

Explanation:

The empirical formular of the compound is obtained using the following steps;

Step 1: Divide the percentage composition by the atomic mass

Hydrogen = 11.21 / 1 = 11.21

Oxygen = 88.79  / 16 = 5.55

Step 2: Divide by the lowest number

Hydrogen  = 11.21 / 5.55 = 2.02 ≈ 2

Oxygen = 5.55 / 5.55 = 1

This means the ratio of the elements is 2 : 1

The empirical formular (simplest formular of a compound) of the compound is;

H₂O

stiv31 [10]3 years ago
3 0

Answer:Empirical formula ======== H₂O    

Explanation:The empirical formula of a compound shows the whole number ratio for  each atom in a compound.

To find empirical formula. we follow the below steps

The total mass of the compound here  is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then  assume 11.21 grams  of hydrogen and 88.79grams of oxygen

                                          Hydrogen                   Oxygen

1.composition by mass    11.21                              88.79

molecular weight              1.007g/mol               15.990g/mol

2.Divide composition by mass  11.21/1.007            88.79/15.99    

by each molecular weight to get 11.13                            5.553

no of moles

3 Divide by the least number of moles

to get atomic ratio                       11.13/5.553          5.553/5.553

                                                         2.004                           1.00

4.Convert  to whole numbers             2                                 1

Empirical formula ======== H₂O    

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2 years ago
Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
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 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
3 years ago
In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g
amm1812
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O 

number of moles = mass / molar mass 
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
3 0
3 years ago
Read 2 more answers
Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
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