Volume mole-mass conversion factors

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

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. 

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. 

A 4.028 m solution of ethylene glycol in water contains 4.028 mol of ethylene glycol per kilogram of water. To find the solution s molarity, we need to find the number of moles of solute per volume (liter) of solution. The volume, in turn, can be found from the mass of the solution by using density as a conversion factor. 

The concentration of a substance in a mixture or solution can be used as a conversion factor to relate the mass (or moles) of a component in a sample of the mixture to the sample volume, or to relate the mass (or molar) flow rate of a component of a continuous stream to the total volumetric flow rate of the stream. Consider, for example, a 0.02-molar solution of NaOH (i.e., a solution containing 0.02 mol NaOH/L) 5 L of this solution contains... 

When reactants are liquids, they are almost always measured by volume. So, to do calculations involving liquids, you add two more steps to the sequence of mass-mass problems—the conversions of volume to mass and of mass to volume. Five conversion factors—two densities, two molar masses, and a mole ratio—are needed for this type of calculation, as shown in Skills Toolkit 4. 

Remember from Chapter 11 that the most convenient unit for counting numbers of atoms or molecules is the mole. One mole contains 6.02 X 10 particles. The molar volume for a gas is the volume that one mole occupies at 0.00°C and 1.00 atm pressure. These conditions of temperature and pressure are known as standard temperature and pressure (STP). Avogadro showed experimentally that one mole of any gas will occupy a volume of 22.4 L at STP. The fact that this value is the same for all gases greatly simplifies many gas law calculations. Because the volume of one mole of a gas at STP is 22.4 L, you can use the following conversion factor to find the number of moles, the mass, and even the number of particles in a gas sample. 

Most chemical reactions that occur on the earth s surface, whether in living organisms or among inorganic substances, take place in aqueous solution. Chemical reactions carried out between substances in solution obey the requirements of stoichiometry discussed in Chapter 2, in the sense that the conservation laws embodied in balanced chemical equations are always in force. But here we must apply these requirements in a slightly different way. Instead of a conversion between masses and number of moles, using the molar mass as a conversion factor, the conversion is now between solution volumes and number of moles, with the concentration as the conversion factor. 

Because the answer we want, molarity, is a ratio of two units (moles of solute—in this case, Na3P04—per liter of solution), we start our unit analysis setup with a ratio of two units. Because we want amount of Na3P04 on the top when we are done, we start with 8.20 g Na3P04 on the top. Because we want volume of solution on the bottom when we are done, we start with 100.0 mL of solution on the bottom. To convert mass of Na3P04 to moles of Na3P04, we use the molar mass of Na3P04. We finish our conversion with a conversion factor that converts milliliters to liters. 

Conversion factors constructed from molarities can be used in stoichiometric calculations in very much the same way conversion factors from molar mass are used. When a substance is pure, its molar mass can be used to convert back and forth between the measurable property of mass and moles. When a substance is in solution, its molarity can be used to convert between the measurable property of volume of solution and moles of solute. 

Between this chapter and Chapter 10, we have now seen three different ways to convert between a measurable property and moles in equation stoichiometry problems. The different paths are summarized in Figure 13.10 in the sample study sheet on the next two pages. For pure liquids and solids, we can convert between mass and moles, using the molar mass as a conversion factor. For gases, we can convert between volume of gas and moles using the methods described above. For solutions, molarity provides a conversion factor that enables us to convert between moles of solute and volume of solution. Equation stoichiometry problems can contain any combination of two of these conversions, such as we see in Example 13.8. 

Thinking it Through It is not a common practice on ACS exams, but do not assume that all information provided in a question is actually essential to its solution. Rather, decide on a route to the solution that will be the most expedient. Given that 6.25 L of Ni(CO)4fgJ form at standard temperature and pressure conditions, the molar volume of 22.4 L can be used as a conversion factor to find the number of moles of Ni(CO>4 present. Then the coefficients in the balanced chemical equation can be applied, observing that one mole of Ni(CO)4 reacts to produce one mole of Ni. The atomic molar mass for nickel can then be used to find the number of grams of Ni present. 

Note you do not need to use the molar mass of Ni(CO)4. Because it is a gas at standard temperamre and pressure conditions, the molar volume is a useful conversion factor to change volume to moles. ... 

This is the amount of heat released when 1 mole of CI2 reacts (see balanced equation). We are not reacting 1 mole of CI2, however. From the volume and density of CI2, we can calculate grams of CI2. Then, using the molar mass of CI2 as a conversion factor, we can calculate moles of CI2. Combining these two calculations into one step, we find moles of CI2 to be ... 

The molar volume, 22.4 L/mol, is used as a conversion factor to convert grams per liter to grams per mole (molar mass) and also to convert liters to moles. The two conversion factors are... 

Molarity has units of mol/L and is therefore a conversion factor between moles and liters. Use the volume and the molarity to find moles of NaCI and then use the molar mass of NaCI (58.5 g/mol) to find grams of NaCI ... 

Also notice that molarity, as expressed in Equations 16.2 and 16.3, includes one unit that is particulate (moles, number of particles) and one unit that is macroscopic (volume, which can be measured). To be practical—to actually work with molarity in the laboratory—you must convert moles to a macroscopic unit, grams. The conversion factor, as you probably know from many uses by now, is molar mass in grams per mole. 

In this problem, molarity is the conversion factor in the second step of the Path, changing the given volume to moles of silver nitrate. Moles are then changed to mass, using molar mass as the third conversion factor. 

All of the solution concentration units introduced in this chapter are direct proportionalities. Percentage concentration by mass is a direct proportionality between mass of solute and mass of solution molarity, between moles of solute and liters of solution molality, between moles of solute and kilograms of solvent and normality, between equivalents of solute and liters of solution. These proportional relationships allow you to think of solution concentration units as conversion factors between the two units in the fraction. Do you know mass of solution and need mass of solute Use percentage concentration. Do you know volume of solution and need moles of solute Use molarity. Thinking about solution concentration units in this way allows you to become more skilled at solving quantitative problems. 

The central focus is again the conversion of a measured quantity to an amount in moles. Because the density is given in g/mL, it will be helpful to convert the measured volume to milliliters. Then, density can be used as a conversion factor to obtain the mass in grams, and the molar mass can then be used to convert mass to amount in moles. Finally, the Avogadro constant can be used to convert the amount in moles to the number of molecules. In summary, the conversion pathway is /xL L g — mol molecules. 

The conversion pathway for this problem is given above. First, convert the volume of the sample to mass this requires density as a conversion factor. Next, convert the mass of halothane to its amount in moles this requires the inverse of the molar mass as a conversion factor. The final conversion factor is based on the formula of halothane. 

Such problems as this one involve many steps or conversions. Try to break the problem into simpler ones involving fewer steps or conversions. It may also help to remember that solving a stoichiometry problem involves three steps (1) converting to moles, (2) converting between moles, and (3) converting from moles. Use molarities and molar masses to carry out volume-mole conversions and gram-mole conversions, respectively, and stoichiometric factors to carry out mole-mole conversions. The stoichiometric factors are constructed from a balanced chemical equation. 

There are circumstances where weight is an important factor (Example 3), but the calculations involving gases may be in terms of volumes of gases involved. The conversion from volumes of gas to mass is done through the numbers of moles. The methods used in these problem solutions are as in Chapter 4 except that the numbers of moles converted to mass (g, lb, etc.) must be determined from the volume, temperature, and pressure of the gases. 

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