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SEP Use Math In another experiment, a chemist combines 7.2 grams of carbon with 19.2 grams of oxygen to form a compound. Identify the compound as either carbon monoxide (CO) or carbon dioxide (CO2).Moles carbon: Moles oxygen: Ratio of the number of carbon atoms to the number of oxygen atoms: Chemical formula for the compound:

SEP Use Math In another experiment, a chemist combines 7.2 grams of carbon with 19.2 grams-example-1

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Step 1 - Remembering the law of mass conservation

As stated by the law of mass conservation, in a chemical process no atoms can be created or destroyed. Consequently, the mass of the system will not change: the sum of the masses of reactants will be exactly equal the sum of the masses of products.

Step 2 - Using the law of mass conservation to understand the problem

The law of mass conservation therefore guarantees that, if we are mixing 7.2 g of C with 19.2 g of O, we will form 7.2+19.2 g of whatever is the product. Therefore we will form 26.4 g of product, which will be either CO or CO2.

Step 3 - Using mass percentage to find the correct product

Since we already know what will the mass of the product be, we can calculate the percentage of C in the product:


\text{ \%(C)=}(7.2)/(26.4)*100=27.3\text{ \%}

Now, let's calculate the percentage of C in both CO (28 g/mol) and CO2 (44 g/mol), remembering that the molar mass of C is 12 g/mol:


\begin{gathered} CO\to\text{ \%(C)=}(12)/(28)*100=42.8\text{ \%} \\ \\ CO_2\to\text{\%(C)=}(12)/(44)*100=27.3\text{ \%} \end{gathered}

We can see that in CO2 the amount percent of C is exactly the same as we have calculated for the unknown product of this reaction. Therefore, the product is CO2.

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