Composition of Mixtures and Pure Substances
You open a bag of trail mix and one scoop is full of raisins while the next has almost none. Now think about a spoonful of table salt: every grain is exactly the same. Why does one kind of sample vary and the other never does? The answer is the heart of this page, and it gives you a tool for finding the formula of an unknown compound.
Pure substances and mixtures
A pure substance has a fixed composition and the same properties throughout. It is either an element (gold, Au) or a compound (water, $\text{H}_2\text{O}$). A mixture is two or more substances physically combined, in any proportions, and each keeps its own identity.
Mixtures come in two types:
-
Homogeneous mixture (solution): uniform at the particle level, so you cannot see separate parts. Example: saltwater, air.
-
Heterogeneous mixture: you can see or separate distinct regions. Example: granite, trail mix, sand in water.
That is why the mass percent of oxygen in water is the same in rain, a lake, or a lab. A mixture has no such rule.
Classify each as a pure substance or a mixture, and if a mixture, as homogeneous or heterogeneous: (a) distilled water, (b) saltwater, (c) granite, (d) gold.
Classify each as a pure substance or a mixture, and if a mixture, as homogeneous or heterogeneous: (a) distilled water, (b) saltwater, (c) granite, (d) gold.
Mass percent of a mixture
To describe how much of one component is in a mixture, use mass percent:
$$\text{mass percent} = \frac{\text{mass of component}}{\text{total mass of mixture}} \times 100\%$$Because a mixture can be made in any proportions, its mass percents can change from sample to sample.
A 50.0 g sample of a trail mix contains 12.5 g of raisins. What is the mass percent of raisins in the mix?
A 50.0 g sample of a trail mix contains 12.5 g of raisins. What is the mass percent of raisins in the mix?
Jayden has a 10.0 g mixture of sand and sodium chloride. He dissolves the salt in water, filters out the sand, and dries it. The dry sand has a mass of 6.4 g. (a) What…
Jayden has a 10.0 g mixture of sand and sodium chloride. He dissolves the salt in water, filters out the sand, and dries it. The dry sand has a mass of 6.4 g. (a) What is the mass percent of NaCl in the original mixture? (b) Explain why the mass percent of NaCl in pure table salt would not vary from sample to sample, while the percent in this mixture can.
Mass percent of a compound
For a compound, the mass percent of an element comes from the formula and molar masses:
$$\text{mass percent of element} = \frac{n \times \text{molar mass of element}}{\text{molar mass of compound}} \times 100\%$$Here $n$ is the number of atoms of that element in one formula unit.
Notice that this answer is the same for any sample of water. That is the law of definite proportions at work.
Calculate the mass percent of oxygen in water, $\text{H}2\text{O}$. Use molar masses of 1.008 g/mol for H and 16.00 g/mol for O.
Calculate the mass percent of oxygen in water, $\text{H}_2\text{O}$. Use molar masses of 1.008 g/mol for H and 16.00 g/mol for O.
From composition to empirical formula
Lab analysis often gives masses or percents, and you want the formula. The empirical formula is the simplest whole-number ratio of atoms in a compound. Moles are the key, because formulas count atoms, not grams.
When you start with masses of the elements (as with the iron oxide sample), skip step 1 and go straight to moles. If only the total and one element are given, find the other element by subtraction.
A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula. Use molar masses of 12.01, 1.008, and…
A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula. Use molar masses of 12.01, 1.008, and 16.00 g/mol.
A 2.500 g sample of a compound of iron and oxygen contains 1.749 g of iron. Determine the empirical formula of the compound. Use molar masses of 55.85 g/mol for Fe and…
A 2.500 g sample of a compound of iron and oxygen contains 1.749 g of iron. Determine the empirical formula of the compound. Use molar masses of 55.85 g/mol for Fe and 16.00 g/mol for O.
Molecular formulas and combustion analysis
The molecular formula is the actual number of atoms in one molecule. It is a whole-number multiple of the empirical formula:
$$\text{molecular formula} = (\text{empirical formula}) \times \frac{\text{molar mass of compound}}{\text{empirical formula mass}}$$For example, glucose is $\text{C}_6\text{H}_{12}\text{O}_6$. Its empirical formula is $\text{CH}_2\text{O}$ (mass 30.03 g/mol), and $180.16/30.03 \approx 6$.
Combustion analysis burns a compound containing C and H in excess oxygen. All carbon ends up as $\text{CO}_2$ and all hydrogen as $\text{H}_2\text{O}$, so you can weigh these products to recover the masses of C and H:
$$m_\text{C} = m_{\text{CO}_2} \times \frac{12.01}{44.01} \qquad m_\text{H} = m_{\text{H}_2\text{O}} \times \frac{2(1.008)}{18.02}$$If the C and H masses add up to less than the sample, the missing mass is another element, usually oxygen.
A hydrocarbon sample is burned completely. A 1.000 g sample produces 3.138 g of $\text{CO}2$ and 1.285 g of $\text{H}2\text{O}$. (a) Calculate the mass of carbon and of…
A hydrocarbon sample is burned completely. A 1.000 g sample produces 3.138 g of $\text{CO}_2$ and 1.285 g of $\text{H}_2\text{O}$. (a) Calculate the mass of carbon and of hydrogen in the sample and confirm that no other element is present. (b) Determine the empirical formula. (c) The molar mass is about 70 g/mol. Determine the molecular formula. Use molar masses of 44.01 g/mol for $\text{CO}_2$, 18.02 g/mol for $\text{H}_2\text{O}$, 12.01 g/mol for C, and 1.008 g/mol for H.
Analyzing a mixture by mass change
You can find the makeup of a mixture when only one component reacts or leaves. Compare masses before and after, and use a balanced equation to connect the lost mass to the component.
The same plan works for the heated $\text{CaCO}_3$ and NaCl mixture: lost mass is the $\text{CO}_2$, and the mole ratio $\text{CaCO}_3 : \text{CO}_2$ is $1:1$.
A 5.00 g sample of a mixture of calcium carbonate, $\text{CaCO}3$, and sodium chloride, NaCl, is heated strongly. Only $\text{CaCO}3$ decomposes:…
A 5.00 g sample of a mixture of calcium carbonate, $\text{CaCO}_3$, and sodium chloride, NaCl, is heated strongly. Only $\text{CaCO}_3$ decomposes: $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$. The solid residue has a mass of 3.90 g. (a) Calculate the mass of $\text{CO}_2$ released. (b) Calculate the mass and mass percent of $\text{CaCO}_3$ in the original mixture. (c) Explain why this method would give wrong results if the NaCl also lost mass on heating. Use molar masses of 100.09 g/mol for $\text{CaCO}_3$ and 44.01 g/mol for $\text{CO}_2$.
Back to the start: a pure substance like table salt has a fixed mass percent of each element (the law of definite proportions), so every grain is the same. Trail mix is a mixture, so the amounts of raisins can differ in every scoop.