Mole-mass conversion factors

A problemsolving flowchart showing the use of mole-mole, mole-mass, mole-volume, and mole-particle conversion factors. 

Just as you cannot make a direct conversion from the mass of jellybeans to the number of jellybeans, you caimot make a direct conversion from the mass of a substance to the number of representative particles in that substance. You must first convert the mass to moles by multiplying by a conversion factor that relates moles and mass. Can you identily the conversion factor The number of moles must then be multiplied by a conversion factor that relates the number of representative particles to moles. That conversion factor is Avogadro s number. 

Solution If we assume that we have 100 g of the compound, then each percentage can be converted directly to grams. In this sample, there will be 40.0 g of C, 6.7 g of H, and 53.3 g of O. Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that... 

This route requires two additional pieces of data the molar mass of the given substance and the mole ratio. The molar mass is determined by using masses from the periodic table. We will follow a procedure much like the one used previously by using the units of the molar mass conversion factor to guide our mathematical operations. Because the known quantity is a mass, the conversion factor will need to be 1 mol divided by molar mass. This conversion factor cancels units of grams and leaves units of moles (see Figure 2.3 below). 

A chemical equation tells us the relations between the amounts (in moles) of each reactant and product. By using the molar masses as conversion factors, we can express these relations in terms of masses. 

Molar mass can be thought of as a conversion factor between mass in grams and number of moles. These conversions are essential in chemistry, because chemists count amounts of substances in moles but routinely... 

Ans. There are 6.02 X 10 i atoms in 1.00 mol Na (Avogadro s number). There is 23.0 g of Na in LOO mol Na (equal to the atomic weight in grams). This problem requires use of two of the most important conversion factors involving moles. Note which one is used with masses and which one is used with numbers of atoms (or molecules of formula units). With numbers of atoms, molecules, or formula units, use Avogadro s number with mass or weight use the formula weight. 

It is possible to use such a conversion factor, but it is advisable while you are learning to use the factors involved with moles to use as few different ones as possible. That way, you have to remember fewer. Also, in each conversion you will change either the unit (mass - moles) or the chemical (CaCO, - O atoms) in a factor, but not both. Many texts do use such combined factors, however. 

If you look at Figure 9-2, you can see that it isn t possible to convert directly between the mass of one substance and the mass of another substance. You must convert to moles and then use the mole-mole conversion factor before converting to the mass of a new substance. The same can be said for conversions from the particles or volume of one substance to that of another substance. The mole is always the intermediary you use for the conversion. 

Solve these kinds of problems by using the definition of molarity and conversion factors. In parts (b) and (c), you must first convert your mass in grams to moles. To do so, you divide by the molar mass from the periodic table (flip to Chapter 7 for details). In addition, be sure you convert milliliters to liters. 

If we want an amount other than 1 mol, we use the molar mass as a conversion factor from the stated number of moles to the mass required ... 

In a mass-to-mass calculation, convert the given mass to moles, apply the mole-to-mole conversion factor to obtain the moles required, and finally convert moles required to mass. 

Then use this molarity as a conversion factor to calculate the number of moles of solute in the stated volume of solution. The mass of solute is given and the number of moles of solute present is now known therefore, to find the molar mass of the solute, divide the mass by the amount. To avoid rounding errors, do the numerical calculation at the end. 

Now that we know how many moles of ethylene we have (0.536 mol), we also know from the balanced equation how many moles of HC1 we need (0.536 mol), and we have to do a mole-to-gram conversion to find the mass of HC1 required. Once again, the conversion is done by calculating the molecular mass of HC1 and using molar mass as a conversion factor ... 

The problem gives the mass of sucrose and asks for a mass-to-mole conversion. Use the molar mass of sucrose as a conversion factor, and set up an equation so that the unwanted unit cancels. 

Use molar mass of Cl2 Use coefficients in as a conversion factor the balanced equation to find mole ratios... 

First, find out how many moles of Cl2 are in 25.0 g of Cl2. This gram-to-mole conversion is done in the usual way, using the molar mass of Cl2 (70.9 g/mol) as the conversion factor ... 

We need to calculate the amount of methyl tert-bu tyl ether that could theoretically be produced from 26.3 g of isobutylene and compare that theoretical amount to the actual amount (32.8 g). As always, stoichiometry problems begin by calculating the molar masses of reactants and products. Coefficients of the balanced equation then tell mole ratios, and molar masses act as conversion factors between moles and masses. 

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