Consider the following data for three binary compounds of...

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \text { \% H (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \hline \text { II } & 12.58 & 87.42 \\ \hline \text { III } & 2.34 & 97.66 \\ \hline \end{array} $$ When 1.00 L of each gaseous compound is decomposed to its elements, the following volumes of $\mathrm{H}_2(\mathrm{~g})$ and $\mathrm{N}_2(\mathrm{~g})$ are obtained: $$ \begin{array}{ccc} & \mathbf{H}_{\mathbf{2}} \mathbf{( L )} & \mathbf{N}_{\mathbf{2}} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \hline \text { II } & 2.00 & 1.00 \\ \hline \text { III } & 0.50 & 1.50 \\ \hline \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

Consider the following data for three binary compounds of hydrogen and nitrogen:
$$
\begin{array}{lcc}
& \text { \% H (by Mass) } & \text { \% N (by Mass) } \\
\hline \text { I } & 17.75 & 82.25 \\
\hline \text { II } & 12.58 & 87.42 \\
\hline \text { III } & 2.34 & 97.66 \\
\hline
\end{array}
$$
When 1.00 L of each gaseous compound is decomposed to its elements, the following volumes of $\mathrm{H}_2(\mathrm{~g})$ and $\mathrm{N}_2(\mathrm{~g})$ are obtained:
$$
\begin{array}{ccc}
& \mathbf{H}_{\mathbf{2}} \mathbf{( L )} & \mathbf{N}_{\mathbf{2}} \text { (L) } \\
\hline \text { I } & 1.50 & 0.50 \\
\hline \text { II } & 2.00 & 1.00 \\
\hline \text { III } & 0.50 & 1.50 \\
\hline
\end{array}
$$
Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

Key Concepts

Determination of Relative Atomic Masses

This concept involves using experimental data, such as percent composition and stoichiometric relationships from chemical reactions, to establish a ratio between the masses of atoms. By comparing how different elements contribute to the overall composition of compounds, one can deduce their relative atomic masses, providing insight into the scale of atomic weights across elements.

Stoichiometric Relationships in Chemical Reactions

Stoichiometry refers to the quantitative relationships between reactants and products in a chemical reaction. By interpreting the volumes of gases produced or consumed (under the same conditions), one can calculate the number of moles involved in the reaction, which is crucial for linking experimental data with theoretical formulas.

Percent Composition Analysis

This concept involves determining the proportion of each element by mass within a compound. It is a fundamental tool in chemistry that allows one to calculate how much of each element is present, based on the percentage values provided, and is essential for formulating the empirical and molecular formulas of substances.

Molecular Formula Determination

Determining a molecular formula requires converting percentage compositions or experimental data into moles of each element, establishing mole ratios, and then using these ratios to deduce both the simplest empirical formula and ultimately the actual molecular formula of a compound. This process ensures that the stoichiometry reflects the true proportions in which the elements combine.

Transcript

00:01 For this problem, we're talking about three different compounds, both contraining hydrogen and nitrogen in different ratios.

00:08 And we're given the percent by mass values for hydrogen and nitrogen for each of the three compounds here.

00:17 So we have hydrogen and nitrogen for one, two, and three as far as their percentage, as well as the volume of the decomposed elements for one liter of this compound.

00:30 Decomposed the volume of the two different elements, h2 and n2.

00:39 And we're trying to use this to figure out what the molecular formulas of the compounds are and also to determine the values for their atomic masses of hydrogen and nitrogen.

00:50 So starting off, if we look over here at the volumes for the decomposed compounds, we can see here that compound 1 has a ratio of 3 to 1, hydrogen to nitrogen.

01:10 Compound 2 has a 2 to 1 ratio, and compound 3 has a 1 to 3 ratio.

01:18 So given that, we can assume that compound 1 with that ratio is going to end up being 1 nitrogen with 3 hydrogen with 3 hydrogen.

01:31 Compounds 2 will be n to h4 which has a 2 to 1 ratio and also makes a complete molecule and then compound 3 will be hn3 because the 3 to 1 ratio now if we figure out the molar masses of this if we assume the atomic mass of hydrogen is 1 .1 1 .008, and the atomic mass of nitrogen is 14 .01, so this will be 14 .01 plus 3 times 1 .008, giving us a value of 17 .034 amu.

02:19 For our second one, it will be 2 times 14 .01 plus 4 times 1 .008, giving us a mass of 32 .052.

02:34 And then finally, for compound three, it'll be three times 14 .01 plus 1 .008, giving a mass of 43 .248.

02:50 And we can check these values by taking the mass and looking at our percentages there.

03:01 So if for compound 1, we have a mass of 17 .034, and we know that the percentage of hydrogen is 17 .75%...

Please give Ace some feedback