An unknown oxide of manganese is reacted with carbon to form...

An unknown oxide of manganese is reacted with carbon to form manganese metal and $\mathrm{CO}_{2}$. Exactly $31.6 \mathrm{~g}$ of the oxide, $\mathrm{Mn}_{x} \mathrm{O}_{\mathrm{r}}$, yielded $13.2 \mathrm{~g}$ of $\mathrm{CO}_{2}$. The simplest formula of the oxide is $(\mathrm{Mn}=55)$ (a) $\mathrm{MnO}$ (b) $\mathrm{Mn} \mathrm{O}_{2}$ (c) $\mathrm{Mn}_{2} \mathrm{O}_{3}$ (d) $\mathrm{Mn} \mathrm{O}$

An unknown oxide of manganese is reacted with carbon to form manganese metal and $\mathrm{CO}_{2}$. Exactly $31.6 \mathrm{~g}$ of the oxide, $\mathrm{Mn}_{x} \mathrm{O}_{\mathrm{r}}$, yielded $13.2 \mathrm{~g}$ of $\mathrm{CO}_{2}$. The simplest formula of the oxide is $(\mathrm{Mn}=55)$
(a) $\mathrm{MnO}$
(b) $\mathrm{Mn} \mathrm{O}_{2}$
(c) $\mathrm{Mn}_{2} \mathrm{O}_{3}$
(d) $\mathrm{Mn} \mathrm{O}$

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Key Concepts

Empirical Formula Determination

This concept involves using experimental mass data to deduce the simplest whole-number ratio of atoms in a compound. It is essential when given the mass of products or reactants from a reaction, allowing chemists to infer the composition and then write the simplest (empirical) formula of an unknown compound.

Stoichiometry

Stoichiometry deals with the quantitative relationships in a balanced chemical equation. It provides the tools to convert between masses, moles, and molecules of reactants and products, ensuring that mass and atoms are conserved in a chemical reaction.

Mole Concept and Molar Mass Calculation

The mole concept allows chemists to relate the mass of a substance to the number of particles it contains using its molar mass. This conversion is fundamental in transforming experimental mass data into moles, which can then be used to determine ratios of atoms in a compound during empirical formula calculations.

Transcript

00:01 First of all, let's have a look what all is given in the problem.

00:05 It is given that the oxide of manganese is reacted with carbon.

00:11 That is, we can say that unknown oxide is reacted.

00:15 So, mnx -o -y plus c will form x -m -n plus co2.

00:23 And it is also given that 31 .6 gram of oxide.

00:28 Mass of mnx, oy is given as 31 .6 grams and it is given that it yielded 13 .2 gram of co2.

00:43 That is mass of co2 is 13 .2 grams.

00:52 And atomic mass of manganese is given as 55.

00:55 What we are supposed to find out, we have to find the formula of the oxide.

01:03 We have to find the formula of the oxide.

01:10 So first of all, write the reaction.

01:15 We know that mnx -o -y reacts with carbon to form x -m -n plus co2.

01:25 Now it is given that 13 .2 gram of co2 is liberated we can calculate number of moles number of moles is nothing but given mass by molar mass molar mass of co2 is 44 given mass is how much 13 .2 grams so we will get it as 0 .3 grams now in this moles of oxygen present will be equal to moles of oxygen will be 0 .3 into 2 because in co2 there are 2 oxygen atom so it will be 0 .6 moles of oxygen of oxygen.

02:04 Now the formula of the oxide is mnx -o -y.

02:08 So it will be x into mn plus y into oxygen equals to 31 .6 gram because this was given that mnx -o -y is 31 .6 gram...

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