For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.įormula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.įinding molar mass starts with units of grams per mole (g/mol). If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. This site explains how to find molar mass. The reason is that the molar mass of the substance affects the conversion. To complete this calculation, you have to know what substance you are trying to convert. In this case, we multiply the molar quantity of the expected NH3 product by the molar mass of NH3 (consult weights on periodic table), as is shown below:Ġ.714 mol NH3 * (17 g/mol NH3) -> 12.138 g NH3įor future problems such as these, just remember the simple workflow that will get you the correct answer for every stoichiometric problem:Ĭreate balanced equation -> convert everything to moles -> determine limiting reagent from what is used up first based upon stoichiometric ratios -> use stoichiometric ratios to determine molar quantity of desired species in relevance to limiting reagent -> convert moles of desired species to requested empirical quantity (mass, pressure, volume, etc).In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.Ī common request on this site is to convert grams to moles. The final step is to leave mole world and go back to gram world. We know from the original balanced equation that it takes 1 mole of N2 to produce 2 moles of NH3, we simply double the amount of N2 being used to determine the amount of NH3 being created. Given that we now know that N2 is the limiting reagent, it is the only reagent that we need to take into consideration further with how much product NH3 is produced. Since we do not have enough N2 to deplete the H2 (only 0.357 moles instead of 0.5), we designate N2 as the limiting reagent driving the reaction in this context, since it will be the reagent used up first. Hence, we have 1.5 mol H2, if we divide by 3, that would mean that it would require 0.5 moles of N2 to deplete all of the H2. When referencing the balanced equation, 3 moles of H2 take 1 mole of N2 to react. Once all of our relevant species are in moles, we must check the limiting reagent by using the stoichiometric values of the species. Simply divide the gram quantities of the reactants by the molar masses to get into "mole world". By referencing the periodic table, we can determine the molar mass of H 2 to be 2 g/mol and the molar mass of N2 to be 28 g/mol. What is the first thing we do? Convert to moles. Now that we have our balanced equation, we can jump directly into stoichiometric conversions. Given that the equation is unbalanced, we must balance it by placing the correct coefficients so that the amount of products on the products side match the reactants on the reactants side as follows: To start, let's logically arrange the reagents and products as presented: N 2 + H 2 -> NH 3 There are three main concepts covered in this question: balancing of equations, stoichiometric conversions, and limiting reagents.
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