Volume 1
1.7. MORE CALCULATIONS WITH ATOMIC MASSES
Mass Percentages of Elements in Compounds • Determination of Atomic Masses of Elements • Exercises
Mass Percentages of Elements in Compounds • Determination of Atomic Masses of Elements • Exercises
1.7.1. Mass Percentages of Elements in Compounds. Using the atomic masses of elements from the periodic table, we can calculate mass percentages of each element in any compound as long as we know its formula. Why is this important?
Imagine running a steel mill. The major component of steel is iron, so you need to purchase iron ores for your business. Your supplier, a mining company, offers two different iron ores for the same price per ton, magnetite (Fe3O4) and hematite (Fe2O3). Suppose that in terms of processing cost, there is no difference between the two. Which of the two ores should you buy? Clearly, you should get the one that has the higher iron content by mass. What is the iron content by mass for Fe3O4 and for Fe2O3? First, let us determine the molecular weights (masses) of the two compounds, using the atomic masses of Fe (56) and O (16).
A molecule of magnetite, Fe3O4, consists of three Fe atoms and four O atoms. Therefore, the mass of the iron atoms in Fe3O4 in a.m.u. is 56 x 3 = 168 and the mass of the oxygen atoms is 16 x 4 = 64. The molecular mass (weight) of magnetite is the sum of the two figures, 168 + 64 = 232. To calculate the mass percentage of iron, we divide the total mass of iron in the molecule by the mass of the molecule (molecular mass) and multiply the result by 100%: (168/232) x 100% = 72.4%. To determine the mass percentage of oxygen in magnetite, we subtract the percentage of iron from 100%.
100% - 72.4% = 27.6%
Alternatively, we can divide the total weight of the oxygen atoms in the molecule by the molecular weight and multiply the result by 100%.
(64/232) x 100% = 27.6%
Regardless of the calculation method, the mass percentage of oxygen in magnetite is 27.6% and the mass percentage of iron is 72.4%.
Next, we repeat the calculation for hematite, Fe2O3. The mass of Fe in the molecule is 56 x 2 = 112 and the mass of O is 16 x 3 = 48. Therefore, the molecular mass of hematite is 112 + 48 = 160. The mass percentages of Fe = (112/160) x 100% = 70% and O = (48/160) x 100% = 30% are calculated the way we just did it for magnetite.
Our calculations show that magnetite contains 72.4% of iron by mass, whereas for hematite the figure is 70%. The difference of 2.4% might look minor. But the steel and chemical industries operate on a huge scale. So now think you are to purchase say 10 million dollars worth of either ore. The difference of 2.4% then translates to $240,000 in savings.
Some readers might say that we do not need to know the atomic masses to find out which of the two ores has the higher iron content. If the formulas are Fe3O4 and Fe2O3, all we need is just simple math to calculate which of the two contains more Fe. In Fe3O4, there are 4/3 O atoms per Fe atom, whereas in Fe2O3, there are 3/2 O atoms per Fe atom. As 4/3 < 3/2, the iron content is higher for Fe3O4 than for Fe2O3. That is true. But, without knowing the atomic masses, we would not be able to calculate the mass percentages of Fe in the ores, which is essential for steel production.
Here is another example. Suppose you want to invest in platinum (Pt). Platinum is more rare and expensive than gold. There is a vendor selling K2PtCl6, K2PtCl4, and Na2PtCl4 for the same price per ounce. The value in this case is determined only by the content of the precious metal, platinum. The other three elements (K, Cl, Na) are vastly cheaper than Pt and their contribution to the value is neglected. We can quickly decide which of the three you should buy by simply looking at the formulas and atomic masses of the elements constituting the compounds.
Let us first compare K2PtCl6 and K2PtCl4. Both molecules contain one platinum (Pt) atom and two potassium (K) atoms. However, K2PtCl6 contains six chlorine (Cl) atoms per one Pt atom, whereas K2PtCl4 contains only four. Therefore, the percentage of Pt by mass is higher for K2PtCl4 and this is the one you should choose between the two.
Next we compare K2PtCl4 and Na2PtCl4. Both contain four chlorine atoms per one platinum atom. However, K2PtCl4 contains two potassium (K) atoms per one Pt atom, whereas Na2PtCl4 contains two sodium (Na) atoms per one Pt atom. From the periodic table we learn that potassium (atomic mass = 39) is heavier than sodium (atomic mass = 23). As the molecular mass of K2PtCl4 is higher than the molecular mass of Na2PtCl4, the platinum is "more diluted" by mass with other elements in K2PtCl4.
The simple considerations above show that the content of Pt by mass diminishes in the order Na2PtCl4 > K2PtCl4 > K2PtCl6. Clearly, if the price per mass unit for the three is the same, you get more Pt for your dollar by purchasing Na2PtCl4.
We can also calculate mass percentages of Pt in each of the three compounds. First, the approximate molecular weights of Na2PtCl4 (383), K2PtCl4 (415), and K2PtCl6 (486) are calculated using the atomic mass values from the periodic table. By dividing the atomic weight of Pt (195) by the calculated molecular weight and then multiplying the result by 100% we get the Pt mass percentage figures, as follows.
(195/383) x 100 % = 51% Pt for Na2PtCl4
(195/415) x 100 % = 47% Pt for K2PtCl4
(195/486) x 100 % = 40% Pt for K2PtCl6.
1.7.2. Determination of Atomic Masses of Elements. Knowing accurate atomic masses of elements is of utmost importance. The 1914 Noble Prize in Chemistry was awarded to the American chemist Theodore W. Richards (1868-1928) for his accurate determinations of the atomic weights of many chemical elements. The first experimental determinations of atomic masses were carried out much earlier though. In the 19th century, John Dalton (England), Thomas Thomson (Scotland), Jöns Jakob Berzelius (Sweden), and Stanislao Cannizzaro (Italy) made critical pioneering contributions to the original determinations of atomic masses. Figure 1-21 displays portraits of these outstanding scientists.
Imagine running a steel mill. The major component of steel is iron, so you need to purchase iron ores for your business. Your supplier, a mining company, offers two different iron ores for the same price per ton, magnetite (Fe3O4) and hematite (Fe2O3). Suppose that in terms of processing cost, there is no difference between the two. Which of the two ores should you buy? Clearly, you should get the one that has the higher iron content by mass. What is the iron content by mass for Fe3O4 and for Fe2O3? First, let us determine the molecular weights (masses) of the two compounds, using the atomic masses of Fe (56) and O (16).
A molecule of magnetite, Fe3O4, consists of three Fe atoms and four O atoms. Therefore, the mass of the iron atoms in Fe3O4 in a.m.u. is 56 x 3 = 168 and the mass of the oxygen atoms is 16 x 4 = 64. The molecular mass (weight) of magnetite is the sum of the two figures, 168 + 64 = 232. To calculate the mass percentage of iron, we divide the total mass of iron in the molecule by the mass of the molecule (molecular mass) and multiply the result by 100%: (168/232) x 100% = 72.4%. To determine the mass percentage of oxygen in magnetite, we subtract the percentage of iron from 100%.
100% - 72.4% = 27.6%
Alternatively, we can divide the total weight of the oxygen atoms in the molecule by the molecular weight and multiply the result by 100%.
(64/232) x 100% = 27.6%
Regardless of the calculation method, the mass percentage of oxygen in magnetite is 27.6% and the mass percentage of iron is 72.4%.
Next, we repeat the calculation for hematite, Fe2O3. The mass of Fe in the molecule is 56 x 2 = 112 and the mass of O is 16 x 3 = 48. Therefore, the molecular mass of hematite is 112 + 48 = 160. The mass percentages of Fe = (112/160) x 100% = 70% and O = (48/160) x 100% = 30% are calculated the way we just did it for magnetite.
Our calculations show that magnetite contains 72.4% of iron by mass, whereas for hematite the figure is 70%. The difference of 2.4% might look minor. But the steel and chemical industries operate on a huge scale. So now think you are to purchase say 10 million dollars worth of either ore. The difference of 2.4% then translates to $240,000 in savings.
Some readers might say that we do not need to know the atomic masses to find out which of the two ores has the higher iron content. If the formulas are Fe3O4 and Fe2O3, all we need is just simple math to calculate which of the two contains more Fe. In Fe3O4, there are 4/3 O atoms per Fe atom, whereas in Fe2O3, there are 3/2 O atoms per Fe atom. As 4/3 < 3/2, the iron content is higher for Fe3O4 than for Fe2O3. That is true. But, without knowing the atomic masses, we would not be able to calculate the mass percentages of Fe in the ores, which is essential for steel production.
Here is another example. Suppose you want to invest in platinum (Pt). Platinum is more rare and expensive than gold. There is a vendor selling K2PtCl6, K2PtCl4, and Na2PtCl4 for the same price per ounce. The value in this case is determined only by the content of the precious metal, platinum. The other three elements (K, Cl, Na) are vastly cheaper than Pt and their contribution to the value is neglected. We can quickly decide which of the three you should buy by simply looking at the formulas and atomic masses of the elements constituting the compounds.
Let us first compare K2PtCl6 and K2PtCl4. Both molecules contain one platinum (Pt) atom and two potassium (K) atoms. However, K2PtCl6 contains six chlorine (Cl) atoms per one Pt atom, whereas K2PtCl4 contains only four. Therefore, the percentage of Pt by mass is higher for K2PtCl4 and this is the one you should choose between the two.
Next we compare K2PtCl4 and Na2PtCl4. Both contain four chlorine atoms per one platinum atom. However, K2PtCl4 contains two potassium (K) atoms per one Pt atom, whereas Na2PtCl4 contains two sodium (Na) atoms per one Pt atom. From the periodic table we learn that potassium (atomic mass = 39) is heavier than sodium (atomic mass = 23). As the molecular mass of K2PtCl4 is higher than the molecular mass of Na2PtCl4, the platinum is "more diluted" by mass with other elements in K2PtCl4.
The simple considerations above show that the content of Pt by mass diminishes in the order Na2PtCl4 > K2PtCl4 > K2PtCl6. Clearly, if the price per mass unit for the three is the same, you get more Pt for your dollar by purchasing Na2PtCl4.
We can also calculate mass percentages of Pt in each of the three compounds. First, the approximate molecular weights of Na2PtCl4 (383), K2PtCl4 (415), and K2PtCl6 (486) are calculated using the atomic mass values from the periodic table. By dividing the atomic weight of Pt (195) by the calculated molecular weight and then multiplying the result by 100% we get the Pt mass percentage figures, as follows.
(195/383) x 100 % = 51% Pt for Na2PtCl4
(195/415) x 100 % = 47% Pt for K2PtCl4
(195/486) x 100 % = 40% Pt for K2PtCl6.
1.7.2. Determination of Atomic Masses of Elements. Knowing accurate atomic masses of elements is of utmost importance. The 1914 Noble Prize in Chemistry was awarded to the American chemist Theodore W. Richards (1868-1928) for his accurate determinations of the atomic weights of many chemical elements. The first experimental determinations of atomic masses were carried out much earlier though. In the 19th century, John Dalton (England), Thomas Thomson (Scotland), Jöns Jakob Berzelius (Sweden), and Stanislao Cannizzaro (Italy) made critical pioneering contributions to the original determinations of atomic masses. Figure 1-21 displays portraits of these outstanding scientists.
Figure 1-21. Left to right: John Dalton (1766-1844) (source), Thomas Thomson (1773-1852) (source), Jöns Jakob Berzelius (1779-1848) (source), and Stanislao Cannizzaro (1826-1910) (source).
Although herein we skip details of how the atomic masses were determined in the original studies, one example below can give you an idea. Knowing the composition of a compound comprising two elements and the atomic mass of one of the two, one can determine the atomic mass of the other element. For example, methane, the main component of natural gas, has the formula CH4. It was determined experimentally that 100 g of methane consists of 25 g of hydrogen and 75 g of carbon. Given the atomic weight of hydrogen (1 a.m.u., originally assigned by John Dalton), we can calculate the atomic weight of carbon. The mass carbon to hydrogen ratio (75 : 25 = 3) shows that one C atom is 3 times heavier than four H atoms. Why four? Because in a molecule of methane, there are four hydrogen atoms per each carbon atom, as indicated by the formula, CH4. Since the mass of four hydrogen atoms equals 4 x 1 a.m.u. = 4 a.m.u., the atomic mass of carbon is 3 x 4 a.m.u. = 12 a.m.u.
1.7.3. Exercises.
1. Calculate mass percentages of each element in the following compounds: (a) CaO, (b) Al2O3, (c) MgBr2, (d) CuSO4, (e) CuSO4•5H2O. Answer
2. You want to invest in gold. There is a vendor selling gold trichloride (AuCl3) and gold tribromide (AuBr3). The price per gram is $100.00 for AuBr3 and $140.00 for AuCl3. Should you buy AuCl3 or AuBr3 to extract more value from your purchase? Neglect the value of the other two elements, Cl and Br. Answer
Although herein we skip details of how the atomic masses were determined in the original studies, one example below can give you an idea. Knowing the composition of a compound comprising two elements and the atomic mass of one of the two, one can determine the atomic mass of the other element. For example, methane, the main component of natural gas, has the formula CH4. It was determined experimentally that 100 g of methane consists of 25 g of hydrogen and 75 g of carbon. Given the atomic weight of hydrogen (1 a.m.u., originally assigned by John Dalton), we can calculate the atomic weight of carbon. The mass carbon to hydrogen ratio (75 : 25 = 3) shows that one C atom is 3 times heavier than four H atoms. Why four? Because in a molecule of methane, there are four hydrogen atoms per each carbon atom, as indicated by the formula, CH4. Since the mass of four hydrogen atoms equals 4 x 1 a.m.u. = 4 a.m.u., the atomic mass of carbon is 3 x 4 a.m.u. = 12 a.m.u.
1.7.3. Exercises.
1. Calculate mass percentages of each element in the following compounds: (a) CaO, (b) Al2O3, (c) MgBr2, (d) CuSO4, (e) CuSO4•5H2O. Answer
2. You want to invest in gold. There is a vendor selling gold trichloride (AuCl3) and gold tribromide (AuBr3). The price per gram is $100.00 for AuBr3 and $140.00 for AuCl3. Should you buy AuCl3 or AuBr3 to extract more value from your purchase? Neglect the value of the other two elements, Cl and Br. Answer