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# Converting Moles to Grams Calculate the mass, in grams, of ISBN: 9780321809247 1

## Solution for problem 1PE Chapter 3.11SE

Chemistry: A Molecular Approach | 3rd Edition

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Problem 1PE

Converting Moles to Grams Calculate the mass, in grams, of 0.433 mol of calcium nitrate. What is the mass, in grams, of (a) 6.33 mol of NaHC03 and (b) 3.0 * 10-5 mol of sulfuric

Step-by-Step Solution:

Answer ; Step 1 In chemistry, the mole is the standard measurement of amount. When substances react, they do so in simple ratios of moles. However, balances give readings in grams. Balances DO NOT give readings in moles. So the problem is that, when we compare amounts of one substance to another using moles, we must convert from grams, since this is the information we get from balances. There are three steps to converting moles of a substance to grams: 1. Determine how many moles are given in the problem. 2. Calculate the molar mass of the substance. 3. Multiply step one by step two. Make sure you have a periodic table and a calculator handy. The three steps above can be expressed in the following proportion: Step 2 , the grams of the substance (upper left) will be the unknown (signified by the letter x). The exact same proportion is used in the grams-to-moles conversion lesson. Then the "x" will reside in the lower left. This proportion is a symbolic equation. When you solve a particular problem, you insert the proper numbers & units into the proper places of the symbolic equation and then you solve using cross-multiplication and division. Also, do not attach units to the unknown. Let it be simply the letter "x." The proper unit should evolve naturally from solving the proportion and cancellation of units. Make sure you have a periodic table and a calculator handy. We have to calculate the mass in grams of 0.0433 moles Calcium nitrate,Ca(NO3)2 . The mass of calcium nitrate present in 0.0433 moles of calcium nitrate is Ca(NO3)2=0.0433moles/164.08g/mol =71.0466g/mol. The formula weight of NaHCO3 is 84.0 amu . So using 1 mol NaHCO3=84.0g NaHCO3. To write the appropriate conversion factor ,we have Grams NaHCO3=(6.33 mol NaHCO3)*(84.0 g NaHCO3/1 mol NaHCO3) =531.72g So , 6.33 mol of NaHCO3=532 g NaHCO3.

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##### ISBN: 9780321809247

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