If the sum is not 100%, you have made an error in your calculations. You can check your work by verifying that the sum of the percentages of all the elements in the compound is 100%: 42.12% + 6.48% + 51.41% = 100.01% The result is shown to two decimal places: mass % C in sucrose = mass of C/mol sucrose molar mass of sucrose × 100 = 144.132 g C 342. The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. Thus 1 mol of sucrose has a mass of 342.297 g note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon. We can then use these masses to calculate the percent composition of sucrose. We can use this information to calculate the mass of each element in 1 mol of sucrose, which will give us the molar mass of sucrose. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. First we will use the molecular formula of sucrose (C 12H 22O 11) to calculate the mass percentage of the component elements then we will show how mass percentages can be used to determine an empirical formula.Īccording to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. For example, sucrose (cane sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. With few exceptions, the percent composition of a chemical compound is constant (see law of definite proportions).-the percentage of each element present in a pure substance-is constant (although we now know there are exceptions to this law). The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass that is, the percent composition The percentage of each element present in a pure substance.
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