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Question: How many grams of \[O\] are contained in \[8.52{\text{ }}g\] \[{K_2}C{O_3}\] ?...

How many grams of OO are contained in 8.52 g8.52{\text{ }}g K2CO3{K_2}C{O_3} ?

Explanation

Solution

Here you should recall the concept of molar mass of any chemical compound which refers to the ratio of mass of a sample of that particular compound and the amount of substance in that particular sample (in moles). The molar mass is generally represented as gmol1gmo{l^{ - 1}} .

Complete step by step answer:
In order to calculate the grams of OO are contained in 8.52 g8.52{\text{ }}g K2CO3{K_2}C{O_3} , first of all we have to calculate the molar mass of K2CO3{K_2}C{O_3} . We know that the molar mass of any compound can be found out by adding the relative atomic masses of each element present in that particular compound. The number of atoms in a compound can be determined from their chemical formula.
Now, let us calculate the molar mass of the given compound i.e. potassium carbonate having a chemical formula of K2CO3{K_2}C{O_3} . We already know the atomic masses of potassium, carbon and oxygen as stated below:
K=39.0983amu C=12.0107amu O=15.999amu  K = 39.0983amu \\\ C = 12.0107amu \\\ O = 15.999amu \\\
Molar mass of this compound can be calculated by adding the mass of two potassium atoms, one carbon atom and three oxygen atoms as shown below:
Molar mass of K2CO3=(2×K)+(1×C)+(3×O) =(2×39.0983)+(1×12.0107)+(3×15.999)=138.2043gmol1  Molar{\text{ }}mass{\text{ }}of{\text{ }}{{\text{K}}_2}C{O_3} = (2 \times K) + (1 \times C) + (3 \times O) \\\ = (2 \times 39.0983) + (1 \times 12.0107) + (3 \times 15.999) = 138.2043gmo{l^{ - 1}} \\\
Now, in order to calculate the number of moles, we generally use the following formula:
Number of moles=Mass(g)Molar mass(gmol1)Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}
Mass of K2CO3{K_2}C{O_3} = 8.52 g8.52{\text{ }}g (Given)
Thus substituting the values, we get:
Number of moles of K2CO3=8.52g138.2043gmol1=0.0616molesNumber{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}{K_2}C{O_3} = \dfrac{{8.52g}}{{138.2043gmo{l^{ - 1}}}} = 0.0616moles
Now, we will use the unitary method to get the final answer.
One mole of K2CO3{K_2}C{O_3} contains 3 moles of oxygen
0.0616 moles0.0616{\text{ }}moles K2CO3{K_2}C{O_3} will contain 31×0.0616=0.1848moles of O\dfrac{3}{1} \times 0.0616 = 0.1848moles{\text{ }}of{\text{ }}O
Mass in grams of oxygen can be calculated from number or moles formula as depicted below:
0.1848=Mass(g)15.999 Mass(g)=15.999×0.1848=2.9566g  0.1848 = \dfrac{{Mass(g)}}{{15.999}} \\\ Mass(g) = 15.999 \times 0.1848 = 2.9566g \\\

Hence, 2.95662.9566 grams of OO are contained in 8.52 g8.52{\text{ }}g K2CO3{K_2}C{O_3} .

Note: Molar mass plays a significant role in chemistry especially during setting up an experiment. During testing principles which involve specific amounts or quantities of a substance, molar mass is used to figure out the exact quantity to be weighed of that particular substance. Basically molar mass is used to determine the stoichiometry in the chemical reactions as well as equations.