We will to use the following conversion factors to solve this problem:
1 mole `~MgCl_2` = molar mass of `~MgCl_2`
1 mole `~MgCl_2` = 1 mole `~Mg^2^+` ions
1 mole `~MgCl_2` = 2 mole `~Cl^1^-` ions
We’ll use the first conversion factor to convert between the given amount in grams of `~MgCl_2` (29.5 g) and moles of `~MgCl_2` .
We’ll use the second and third conversion factors to convert between moles of `~MgCl_2` and moles of of `~Mg^2^+ ` ions and `~Cl^1^-` ions.
Step 1: Determine the molar mass of `~MgCl_2` . We need to do this calculation first, because we will need to use the molar mass of `~MgCl_2` as part of our first conversion factor.
Calculate the molar mass by: 1) multiplying the atomic mass of each element in the compound times its subscript, 2) adding your answers together.
Molar Mass of `~MgCl_2` :
Mg: 1 x 24.305 = 24.305 g
Cl: 2 x 35.453 = 70.906 g
24.305 g Mg + 70.906 g Cl = 95.211 g `~MgCl_2`
So, our first conversion factor becomes:
1 mole `~MgCl_2` = 95.211 g `~MgCl_2`
Step 2: Convert grams of `~MgCl_2` to moles `~MgCl_2` .
To convert grams to moles, multiply the given amount of `~MgCl_2` times the first conversion factor shown above.
(29.5 g `~MgCl_2` )(1 mole `~MgCl_2` /95.211 g `~MgCl_2` ) = 0.310 mole `~MgCl_2`
Notice that the conversion factor is oriented such that the part of the conversion factor with the same unit as the given unit is placed in the denominator. This enables you to cancel out the given unit and be left with the unit that you are trying to convert to.
Step 3: Convert moles of `~MgCl_2` to moles of `~Mg^2^+` ions and moles of `~Cl^1^-` ions.
To convert moles of `~MgCl_2` to moles of ions, we need to multiply moles of `~MgCl_2` (0.310 moles) by the second and third conversion factors shown above.
Moles of `~Mg^2^+` ions:
(0.310 mol `~MgCl_2` )(1 mol `~Mg^2^+` /1 mol `~MgCl_2` ) = 0.310 mol `~Mg^2^+` ions
Moles of `~Cl^1^-` ions:
(0.310 mol `~MgCl_2` )(2 mol `~Cl^1^-` /1 mol `~MgCl_2` ) = 0.620 mol `~Cl^1^- ` ions
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