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Now that we know how to convert from atoms to molecules, from molecules to moles and from moles to grams we can string these conversion factors together to solve more complicated problems.
Example \(\PageIndex{6}\): Converting between grams and Atoms How many atoms of hydrogen are in 4.6 g of CH3OH? Solution First we need to determine the mass of one mole of methane (CH3OH). Using the periodic table to find the mass for each mole of our elements we have: \[1\, mole\, C \,= 1\, \cancel{mole\, C}\,\times \left(\frac{12.011\, g\, C}{1\,\cancel{mole\,C}} \right)\, = 12.011 \, g\, C\] \[4\, mole\, H \,= 4\, \cancel{mole\, H}\,\times \left(\frac{1.008\, g\, H}{1\,\cancel{mole\,H}} \right)\, = 4.032 \,g\, H\] \[1\, mole\, O \,= 1\, \cancel{mole\, O}\,\times \left(\frac{15.999\, g\, O}{1\,\cancel{mole\,O}} \right)\, = 15.999 \, g\,O\] Adding the masses of our individual elements have:\[12.011\, g \,+ \,4.032\,g\, +\, 15.999\, g\, =\,32.042\, g\, CH_{3}OH\] As we were calculating the molar mass of CH3OH we have \[32.042\, g\, CH_{3}OH\, = \,1\, mole \,CH_{3}OH\] Which we can use as a conversion factor \[\frac{32.042\, g\, CH_{3}OH}{1\, mole\, CH_{3}OH}\, or\, \frac{1\, mole\, CH_{3}OH}{32.042\, g\, CH_{3}OH}\] We also know that for every molecule of CH3OH we have 4 atoms of H. Now we can go back to the starting value given in the question: \[4.6\,\cancel{g\, CH_{3}OH}\,\times\,\left(\frac{1\, \cancel{mole\, CH_{3}OH}}{32.042\,\cancel{g\,CH_{3}OH}}\right)\times\left(\frac{6.022\, x\, 10^{23}\,\cancel{\,molecules\, CH_{3}OH}}{1 \,\cancel{mole\, CH_{3}OH}}\right)\times\left(\frac{4\, atoms\, H}{1 \cancel{molecule \,CH_{3}OH}}\right) = 3.5\,x\,10^{23}\, atoms\, H\]
Example \(\PageIndex{7}\): Converting from Grams to grams How many grams of oxygen are in 3.45 g of H3PO4 Solution First we need to determine the mass of one mole of phosphoric acid \(\ce{H_{3}PO_{4}}\) We have: \[3\, mole\, H \,= 3\, \cancel{mole\, H}\,\times \left(\frac{1.008\, g\, H}{1\,\cancel{mole\,H}} \right)\, = 3.024\,g\, H\] \[1\, mole\, P \,= 1\, \cancel{mole\, P}\,\times \left(\frac{30.974\, g\, P}{1\,\cancel{mole\,P}} \right)\, = 30.974 \,g\, P\] \[4\, mole\, O \,= 4\, \cancel{mole\, O}\,\times \left(\frac{15.999\, g\, O}{1\,\cancel{mole\,O}} \right)\, = 63.996 \,g\, O\] Adding the masses of our individual elements have:\[3.024\, g \,+ \,30.974\,g\, +\, 63.996\, g\, =\,97.994\, g\, H_{3}PO_{4}\] As we were calculating the molar mass of CH3OH we have \[97.994\, g\, H_{3}PO_{4}\, = \,1\, mole \,H_{3}PO_{4}\] Which we can use as a conversion factor \[\frac{97.994\, g\, H_{3}PO_{4}}{1\, mole\, H_{3}PO_{4}}\, or\, \frac{1\, mole\, H_{3}PO_{4}}{97.994\, g\, H_{3}PO_{4}}\] To calculate the molar mass of \(\ce{H_{3}PO_{4}})\ we used the idea that we have 3 moles of H, 1 mole of P and 4 moles of O for every one mole of \(\ce{H_{3}PO_{4}}). We recall that we can make our own conversion factors as long as the top and bottom are equal to each other. So much the same way I can give you four quarter or 1 dollar, if I hand you a mole of \(\ce{H_{3}PO_{4}}) I've handed you 3 mole of H, 1 mole of P and 4 moles of O. As this is true there is an additional set of conversion factors we can use: \[\left(\frac{3\, moles\, H}{1\, mole\, H_{3}PO_{4}}\right)\, and \, \left(\frac{1\, mole\, P}{1\, mole\, H_{3}PO_{4}}\right)\, and\, \left(\frac{4\,moles\, O}{1\, mole\, H_{3}PO_{4}}\right)\] Now we can go back to the starting value given in the question: \[3.45\,\cancel{g\,H_{3}PO_{4}}\,\times\,\left(\frac{1\,\cancel{mole\,H_{3}PO_{4}}}{97.994\,\cancel{g\,H_{3}PO_{4}}}\right)\times\left(\frac{4\,\cancel{mole\,O}}{1\,\cancel{\,mole\, H_{3}PO_{4}}}\right)\times\left(\frac{15.999\, g\, O}{1 \cancel{mole \,O}}\right) = 2.25\, g\, O\]
Molar mass of Ca = 40.078 g/mol Convert grams Ca to moles or moles Ca to grams ›› Percent composition by element
›› Calculate the molecular weight of a chemical compoundIn 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. A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. 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. 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. Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
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