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Vaselesa [24]
3 years ago
11

Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?

Chemistry
1 answer:
tester [92]3 years ago
5 0

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the <em>empirical formula </em>

EF = K₂O

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If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?
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The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃  = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

                               CaCO₃ (s)   ⇒   CaO (s)   +  CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

                    Mass of reactants = Mass of products

                    Mass of CaCO₃   = Mass of CaO + Mass of CO₂

Solve for CO₂

                    Mass of CO₂  = Mass of CaCO₃ - Mass of CaO                    

                     Mass of CO₂ = 983 g - 551 g

Simplification

                     Mass of CO₂ = 432 g                        

         

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