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Avogadro's number is the number 6. 0221367 x 10 ^ 23, commonly rounded to just three significant digits: 6. 02 x 10 ^ 23, and is the number of representative particles in a mole. Avogadro's number is commonly used to compute the quantities of substances involved in chemical reactions, called stoichiometry, and is one of the most important and versatile components of modern chemistry. Avogadro's number is named after the Italian physicist Amadeo Avogadro. Avogadro's number first came about when Amadeo Avogadro proposed Avogadro's law in 1811.
Avogadro's law states that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal amount of molecules. The specific number of molecules in one gram-mole of a substance is the value 6. 02 x 10 ^ 23. For example, the molecular weight of oxygen is 32, so one gram-mole of oxygen has a mass of 32 grams and contains 6. 02 x 10 ^ 23 molecules (Black). Avogadro's number also deals with the mole. The mole is a SI unit used to measure the amount of a substance, and is equal to Avogadro's number. It is equal to the number of carbon atoms in exactly 12 grams of carbon- 12.
A mole of any substance contains 6. 022127 x 10 ^ 23 representative particles. A representative particle is any type of particle, such as atoms, molecules, formula units, electrons or ions, and 1 mole of any substance always contains the same number of molecules. Avogadro's number relates the mass of a mole of a substance to the mass of a single molecule. For example, to find the mass of one molecule of H 2 O, you would use the formula: (Mass) / (Avogadro's number) Since the molecular mass of H 2 O is 18, then to formula would be: (18 g) / (6. 02 x 10 ^ 23) By using this formula you discover that the mass of one molecule 2. 99 x 10 ^- 23 grams (Dickson 106). A major use of the mole is stoichiometry.
The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and method (meaning "measure"). Stoichiometry deals with calculations of the masses of reactants and products in a chemical reaction, and is a very mathematical part of chemistry. You can use stoichiometry to calculate moles, masses, and percent composition within a chemical equation (Burns 312). In solving stoichiometry problems, the molar ratio is very important. These rations are the coefficients of a balanced equation.
For example, given the equation: 2 H 2 + O 2 -- -> 2 H 2 O The molar ratio between H 2 and O 2 is 2: 1, and the molar ratio between O 2 and H 2 O is 1: 2. There are many types of problems involving molar ratios, and one type is a mole-mole problem (Park). An example of this type is: Given the equation N 2 + 3 H 2 -- -> 2 NH 3 Find the number of moles of NH 3 that are produced when 2 moles of N 2 react with a sufficient amount of H 2. To solve this, you form the equation: 2 = X 1 2 Cross-multiply and you find out that 4 moles of NH 3 are produced. Another type of problem is a mole-mass problem. An example of this type is: Given the equation 2 K ClO 3 -- -> 2 KCl + 3 O 2 Find how many grams of O 2 are produced when 1. 5 moles of K ClO 3 decomposes.
You solve this by setting up a proportion of molar ratios. The ratio of K ClO 3: O 2 is 2: 3, and the ratio from the problem is 1. 5: X. The proportion is then: 2 = 1. 5 3 X Cross-multiply and you find that 2. 25 moles of O 2 is produced. Since the molar mass of O 2 is 32, than 2. 25 x 32 gives you 72 grams of O 2 produced. These types of problems can work in reverse.
For example, using the same equation, you could figure out how many moles of K ClO 3 decompose when 80. 0 grams of O 2 are produced. The ratio of O 2: K ClO 3 is 3: 2, and the ratio from the problem is 2. 5: X, since 80 g 32. 0 g = 2. 5 moles. The proportion is then: 3 = 2. 5 2 X Cross-multiply and you find that 1. 67 moles of K ClO 3 decomposed. The most common type of problem is a mass-mass problem. An example of this is: Given the reaction 2 Au Cl 3 -- -> 2 Au + 3 Cl 2 Find how many grams of chlorine can be produced from the decomposition of 64 g of Au Cl 3. First you find the number of moles: 64 g = . 211 moles 303. 32 g / mole (mass of Au Cl 3) Next, you set up the proportion: . 211 = 2 X 3 Cross-multiply and you get. 316 moles.
Multiply that by the mass of Cl 2, 70. 906 g, and you find out that 22. 4 g of chlorine are produced from the decomposition of 64 g of Au Cl 3 (Park). A final type of problem using the mole is percent composition problems. The basic formula used in these types of problems is: percent by mass = mass of part mass of whole There are two types of percent composition problems. One is problems in which you are given the formula, or the weight of each part, and asked to calculate the percentage of each element and the other is problems in which you are given the percentages and asked to calculate the formula. In percent composition problems it is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as much as possible, this is called the empirical formula. An example of this question is: If a compound is 47. 3 % carbon, 10. 6 % hydrogen and 42. 0 % sulfur, what is its empirical formula?
To do this problem you need to transfer all the percents to masses. You assume that there is 100 g of this substance. Then you convert to moles: Next you try to get an even ratio between the elements. To do this you divide by the number of moles of sulfur, since it is the smallest number: The resulting formula is then Carbon 3 Hydrogen 8 Sulfur.
An example of a problem in which you need to calculate the percent of each element is: What is the percent by mass of hydrogen sulfate, H 2 SO 4? To do this you first need to calculate the total weight of the compound, which comes out to be 98. 09 g. Next you take the weight of each element over the total mass and multiply by 100 to get a percent: After doing this, you find out that that H 2 SO 4 is made up of 2. 06 % hydrogen, 32. 7 % sulfur, and 65. 2 % oxygen by mass (Stoichiometry). As you can see, Avogadro's number is a number equal to a mole, and is used in stoichiometry. This number can be used in many different situations, whether it is finding the mass of a molecule, mole-mole problems, mole-mass problems, mass-mass problems, and percent composition problems. Avogadro's number is used in all these types of problems, and is one of the most resourceful and multipurpose factors of modern chemistry.
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