When a gaseous compound X containing only C, H, and O is...

When a gaseous compound $X$ containing only $C, H,$ and $O$ is burned in $O_{2}, 1$ volume of the unknown gas reacts with 3 volumes of $O_{2}$ to give 2 volumes of $C O_{2}$ and 3 volumes of gaseous $\mathrm{H}_{2} \mathrm{O}$ . Assume all volumes are measured at the same temperature and pressure. (a) Calculate a formula for the unknown gas, and write a balanced equation for the combustion reaction. (b) Is the formula you calculated an empirical formula or a molecular formula? Explain. (c) Draw two different possible electron-dot structures for the compound $\mathbf{X}$ . (d) Combustion of 5.000 $\mathrm{g}$ of $\mathrm{X}$ releases 144.2 $\mathrm{kJ}$ heat. Look up $\Delta H_{\mathrm{f} \text { values for }} \mathrm{CO}_{2}(g)$ and $\mathrm{H}_{2} \mathrm{O}(g)$ in Appendix $\mathrm{B}$ , and calculate $\Delta H_{\mathrm{f}}^{\circ}$ for compound $\mathrm{X}$ .

When a gaseous compound $X$ containing only $C, H,$ and $O$ is burned in $O_{2}, 1$ volume of the unknown gas reacts with 3 volumes of $O_{2}$ to give 2 volumes of $C O_{2}$ and 3 volumes of
gaseous $\mathrm{H}_{2} \mathrm{O}$ . Assume all volumes are measured at the same temperature and pressure.
(a) Calculate a formula for the unknown gas, and write a balanced equation for the combustion reaction.
(b) Is the formula you calculated an empirical formula or a molecular formula? Explain.
(c) Draw two different possible electron-dot structures for the compound $\mathbf{X}$ .
(d) Combustion of 5.000 $\mathrm{g}$ of $\mathrm{X}$ releases 144.2 $\mathrm{kJ}$ heat. Look
up $\Delta H_{\mathrm{f} \text { values for }} \mathrm{CO}_{2}(g)$ and $\mathrm{H}_{2} \mathrm{O}(g)$ in Appendix $\mathrm{B}$ , and calculate $\Delta H_{\mathrm{f}}^{\circ}$ for compound $\mathrm{X}$ .

Key Concepts

Gas Stoichiometry and Avogadro's Law

This concept involves using the idea that equal volumes of gases, under the same temperature and pressure, contain an equal number of molecules, allowing for volume ratios to be directly used as mole ratios in chemical reactions. This is central to determining the proportions in which gases react or are produced in a reaction.

Combustion Reactions

Combustion reactions are chemical reactions in which a substance reacts with oxygen to yield combustion products, typically carbon dioxide and water when organic compounds are involved. These reactions require balancing the equation to ensure that the number of atoms of each element is the same on both sides of the reaction.

Empirical and Molecular Formulas

The empirical formula represents the simplest whole-number ratio of the elements in a compound, while the molecular formula shows the actual number of atoms of each element in a molecule. Distinguishing between the two is important for understanding the composition and structure of chemical compounds.

Lewis Structures (Electron-Dot Diagrams)

Lewis structures are diagrams that represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist. They are essential for visualizing the arrangement of electrons, predicting molecular geometry, and understanding the reactivity and properties of a compound.

Enthalpy of Formation and Hess's Law

This concept involves calculating the heat change during reactions, particularly combustion, using standard enthalpies of formation. Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken, enabling the determination of unknown enthalpy values by combining known values.

Transcript

00:01 In order to find our unknown compound, we're going to have to use a little bit of math to understand the relationship between our various elements.

00:10 And we are going to be using a very simple systems of equation to understand how we would find the moles of our compound.

00:20 So let's first note that our compound x is containing carbon, hydrogen, and oxygen.

00:27 And we're told we have this reaction where we have x plus three diatomic oxygen that's going to be forming two moles of carbon dioxide plus three moles of water.

00:39 Now this is where it gets a little bit confusing mathematically.

00:43 So we are going to have to find the number of atoms of each carbon, hydrogen, and oxygen.

00:52 And this is the reaction we're going to use to do that.

00:54 We know that c subx, h sub y, o sub z, plus x plus y over 4, minus z over 2, forms x times carbon dioxide plus y over 2 times water.

01:09 This seems a little bit complicated, but all these variables are saying is that we have coefficients in front of our products and reactants, and we have a number that denotes a number of atoms in our compound.

01:21 Now from this equation we can deduce that x equals 2 and y equals 6.

01:29 This was found by simply comparing this equation to the one above.

01:34 And now we need to find z.

01:36 So we'll get x plus y over 4 minus z over 2 is equivalent to 3 and what this means is when we simplify it z is equivalent to 1.

01:46 So that means our formula is c2 h6 o.

01:51 Now the way we know that this is a molecular formula, not an empirical formula, it's because remember we're dealing with diatomic oxygen.

02:01 And because oxygen is diatomic, we could have various empirical formulas...