Let's find Molar mass of [tex] \bf CO_2[/tex]
[tex]\qquad[/tex][tex] \pink{\bf\longrightarrow { Molar \:mass \:of \: CO_2:-} }[/tex]
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow 12 + 16 \times 2 [/tex]
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow 44[/tex]
[tex]\qquad[/tex]_____
To calculate the percentage of O in [tex] \bf CO_2[/tex], we have find out the ratio of 'Molar mass of O and molar mass of [tex] \bf CO_2[/tex].'
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow \dfrac{Molar\: Mass_{(O)}}{Molar \: Mass _{(CO_2)} }[/tex]
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow \dfrac{16 }{ 44}[/tex]
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow \cancel{\dfrac{16 }{ 44}}[/tex]
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow 0.3636[/tex]
Now, multiply this value by 100 ℅.
[tex]\qquad[/tex][tex] \bf \twoheadrightarrow 0.3636 \times 100 \%[/tex]
[tex]\qquad[/tex][tex] \pink{\bf \twoheadrightarrow 36.36 \%} [/tex]
Considering the definition of molar mass, the percent of O in CO₂ is 36.35%.
The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.
In this case, you know the molar mass of the elements is:
So, the molar mass of the compound CO₂ is calculated as:
CO₂= 12.01 amu + 2× 16 amu
Solving:
CO₂= 44.01 amu
In first place, you can observe that 1 mole of carbon dioxide contains 1 mole of O.
Oxygen has a molar mass of 16.00 amu, so 1 mole of carbon atoms has a mass of 16.00 grams.
This means that if you take 44.01 g of carbon dioxide, it will contain:
[tex]\frac{Molar Mass_{O} }{Molar mass_{CO_{2} } } = \frac{16 amu}{44.01 amu} =0.3635[/tex]
Expressing as a percentage:
0.3636= 36.35%
The percent of O in CO₂ is 36.35%.
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