Avogadros 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. Avogadros 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. Avogadros number is named after the Italian physicist Amadeo Avogadro. Avogadros number first came about when Amadeo Avogadro proposed Avogadros law in 1811. Avogadros 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 (Blauch).
Avogadros number also deals with the mole. The mole is a SI unit used to measure the amount of a substance, and is equal to Avogadros 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. Avogadros number relates the mass of a mole of a substance to the mass of a single molecule.
The Term Paper on The Number of Moles of Water
Many salts that have been crystallized from water solutions appear to be perfectly dry, yet when heated they discharge large quantities of water. An example can be hydrated copper (II) sulfate. Exactly this salt is used in the described experiment. RESEARCH QUESTION What is the number of moles of water of crystallization associated with one mole of copper (II) sulfate, in the hydrate CuSO4 * xH2O ...
For example, to find the mass of one molecule of H2O, you would use the formula: ( Mass ) / (Avogadros number ) Since the molecular mass of H2O is 18, then to formula would be: ( 18g ) / ( 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 metron (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 H2 + O2 —> 2 H2O The molar ratio between H2 and O2 is 2:1, and the molar ratio between O2 and H2O 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 N2 + 3 H2 —> 2 NH3 Find the number of moles of NH3 that are produced when 2 moles of N2 react with a sufficient amount of H2. To solve this, you form the equation: 2 = X 1 2 Cross-multiply and you find out that 4 moles of NH3 are produced. Another type of problem is a mole-mass problem. An example of this type is: Given the equation 2 KClO3 —> 2 KCl + 3 O2 Find how many grams of O2 are produced when 1.5 moles of KClO3 decomposes.
You solve this by setting up a proportion of molar ratios. The ratio of KClO3 : O2 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 O2 is produced. Since the molar mass of O2 is 32, than 2.25 x 32 gives you 72 grams of O2 produced. These types of problems can work in reverse. For example, using the same equation, you could figure out how many moles of KClO3 decompose when 80.0 grams of O2 are produced. The ratio of O2 : KClO3 is 3:2, and the ratio from the problem is 2.5 : X, since 80g 32.0g = 2.5 moles.
The Term Paper on Solving Problems and Making Decisions 2
Background As the Head Phlebotomist at the RD&E Wonford site I oversee the day to day running and supervision of 32 members of staff. 20 staff members are contracted, working between 12 and 37.5 hours per week and the remainder are bank workers working on an ad-hoc basis when required. Daily we have a minimum of 15 phlebotomists working throughout the site. The role of the phlebotomist is ...
The proportion is then: 3 = 2.5 2 X Cross-multiply and you find that 1.67 moles of KClO3 decomposed. The most common type of problem is a mass-mass problem. An example of this is: Given the reaction 2 AuCl3 —> 2 Au + 3 Cl2 Find how many grams of chlorine can be produced from the decomposition of 64g of AuCl3. First you find the number of moles: 64g = .211 moles 303.32 g/mole (mass of AuCl3) Next, you set up the proportion: .211 = 2 X 3 Cross-multiply and you get .316 moles. Multiply that by the mass of Cl2, 70.906g, and you find out that 22.4g of chlorine are produced from the decomposition of 64g of AuCl3 (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 Carbon3 Hydrogen8 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, H2SO4? To do this you first need to calculate the total weight of the compound, which comes out to be 98.09g. 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 H2SO4 is made up of 2.06% hydrogen, 32.7% sulfur, and 65.2% oxygen by mass (Stoichiometry).
The Essay on Mass Determination
Objective: The purpose of this experiment is to see the difference of precision of different balances. When doing experiments we determine the mass my measuring the sample with a balance. There are many kinds of balances that measure to different precisions. This experiment shows the different results that two balances can give. Summary of Procedures Determine the mass with the triple beam scale ...
As you can see, Avogadros 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. Avogadros number is used in all these types of problems, and is one of the most resourceful and multipurpose factors of modern chemistry..
