A compound X contains 63.3 percent manganese (Mn) and 36.7...

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When $X$ is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent $\mathrm{Mn}$ and 28.0 percent $\mathrm{O}$ is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of $\mathrm{X}$ to $\mathrm{Y}$

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When $X$ is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent $\mathrm{Mn}$ and 28.0 percent $\mathrm{O}$ is formed.
(a) Determine the empirical formulas of X and Y.
(b) Write a balanced equation for the conversion of $\mathrm{X}$ to $\mathrm{Y}$

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is the method of ensuring that the number of atoms of each element is the same on both sides of the reaction. This guarantees that mass and charge are conserved throughout the chemical process, and it is crucial for accurately describing reactions, such as the conversion of one compound to another.

Percent Composition

Percent composition provides the mass percentage of each element in a compound. This concept is fundamental for determining the relative amounts of elements, which is a necessary first step in deducing the compound’s empirical formula.

Empirical Formula Determination

Empirical formula determination involves converting percentage compositions to moles of each element using their atomic masses, then finding the simplest whole-number ratio of the moles. This process reveals the simplest representative formula of the compound.

Stoichiometry of Decomposition Reactions

Stoichiometry in the context of decomposition reactions involves understanding how compounds break down into simpler substances or different compounds, and it requires balancing the reaction based on the conservation of mass for each element. This is particularly relevant when evaluating how the empirical formula changes upon decomposition.

Transcript

00:03 And we have given information for part a, compound number of eggs and percent of manganese by mass is 63 .3%.

00:48 The percent of oxygen is 36 .7 percent.

01:02 Let us assume that the mass of a compound is 100 grams.

01:28 Hence, the mass of magnesium oxygen is.

01:41 Manganese and oxygen is mass of manganese is equal to 63 .3 grams.

02:02 And mass of oxygen is equal to 36 .7 gram.

02:13 Here first we will find the number of moles of manganese and oxygen.

02:37 So number of moles of n -mini manganese is equal to mass of manganese devalued.

02:49 Divided by molar mass of manganese is equal to mass of manganese is 63 .3 gram divided by molar mass of manganese is 54 .983 .938 per mole is equal to the calculated value is 1 .938.

03:27 0 .152 mole and number of moles of oxygen is equal to mass of oxygen divided by molar mass of oxygen is equal to mass of oxygen is 36 .7 gram divided by molar mass of oxygen is 15 .99 gram per mole.

04:08 Is equal to gram gram cancel and we have 2 .294 mole.

04:28 And to find the empirical formula, empirical formula, we have to divide each number of moles by the smallest one.

04:58 So we have the number of mingney's atoms is equal to 1 .152 mall divided by 1 .152 mall is the smallest number of mall is equal to 1 .152.

05:59 Oxygen atoms is equal to 2 .294 moles divided by sumulus number of more is 1 .152 mole is equal to 2.

06:30 Hence the empirical formula of x is mn .02.

07:05 The next step is compound number y by y...

Please give Ace some feedback