How many grams of Green #3 dye (808.84 g/mol) are needed to prepare 5.00 L of a 3.08 x 10^-5 M solution?

Answer 1

To start off with, we should calculate how many moles of green dye are in our 5 liter solution. This will require that we work backwards somewhat from our finished product.

We want 5 L of 3.08 x 10^-5 M solution of green #3 dye. Molarity is expressed in the units of moles/liter, so we are saying that we want 5 liters of 3.08 x 10^-5 mol/L solution. To determine the number of total moles needed, we simply need to multiply the number of liters we would like to make by the concentration of our desired solution. Doing so will cancel out the units of L and leave us with moles.

5 L x (3.08 x 10^-5 mol/L) = ‭0.000154‬ or 1.54 x 10^-4 moles of green #3 dye

Next, we simply need to determine how many grams of green dye there are in 1.54 x 10^-4 moles. This is done using the molar mass given in the question. Again, we can multiply these because it will result in the units for moles being cancelled out, leaving us with grams.

1.54 x 10^-4 mol x 808.84 g/mol = 0.125 grams of green #3 dye

Answer: 0.125 grams of green #3 dye are needed to prepare 5.00 L of a 3.08 x 10^-5 M solution.

Answer 2

Hi Jen,

Now, come on, you can figure this one out just from the units on the data (that's called "dimensional analysis", just so you know). You want to get g (grams) out, right? For this problem, you are not adding or subtracting any materials, so it must be a multiplication- and division- type expression. The only way of combining the data to do that is volume x concentration x molecular weight; that's (as units) L x (mol/L) x (g/mol) = g . That's why, when you try to solve any problem involving values with units, you always write the units down with every number. Then you calculate the numbers, and figure (by cancelling) the units. If you don't come out with exactly the units you wanted, you either aren't finished, or you set up the calculation wrong (divided vs. multiplied, etc.).

Now a note as to how to set up calculations efficiently: start with a blank expression. Take your first datum (volume), and ask yourself: if I have a greater volume of my liquid, would that give me: more grams of analyte (Green #3 dye) [:substitute in whatever the target quantity is, if not grams of analyte] ? Intuitively, you should be able to say, "yes!". That means, use this datum (value and units!) in the numerator of the expression you are building. If instead the answer is "no, that would give me less (grams of analyte)", then you would put it in the denominator. Once you have built such an expression, it's usually the correct one (but, always check that the units come out right!).

You should practice thinking along these lines, as you set up expressions, until it's second nature to you.

Cheers, --Mr. d.