I need help with a concentration question. I need to know how to calculate the product with the higher concentration. $6.50 for 185 mL and the concentration of the active ingredient is 0.005% OR $7.00 for 165mL and the concentration of the active ingredient is 0.700%.

I need to know how to write this out so I can solve others like it..If anyone can help, it would be greatly appreciated.

Your question isn't quite clear. If you really DO want the product with the higher concn, it MUST be the 0.700%. Obviously 0.700% is a higher concn than 0.005% and there is no calculation to it. However, I suspect you really want the price per gram of each. Here is how you do that (and I'm assuming the concn quote is percent w/v and not w/w.

grams active ingredient in the first one is
(0.005 g/100 mL) x 185 mL = 0.00925 g.
That costs $6.50 so the price per gram is $6.50/0.00925 g = $702.70/gram.

The price of the other one per gram can be done exactly the same way. I obtained about $6.00/gram (but that isn't the exact answer) so the first one is much more expensive than the latter.

Thank you Dr. Bob. You were correct in your assumption. I had been staring at that problem for too long...It's very simple now that I see your solution. Thanks again for your help!

To calculate the product with the higher concentration, you need to compare the amount of the active ingredient (in grams) contained in each product.

Step 1: Convert the concentrations to decimal form. To do this, divide the percentages by 100.

For the first product: 0.005% = 0.005/100 = 0.00005
For the second product: 0.700% = 0.700/100 = 0.007

Step 2: Calculate the amount of the active ingredient (in grams) in each product by multiplying the concentration by the volume in milliliters.

For the first product: 0.00005 * 185 = 0.00925 grams
For the second product: 0.007 * 165 = 1.155 grams

Step 3: Compare the amounts of the active ingredient.
Since 1.155 grams is greater than 0.00925 grams, the second product ($7.00 for 165mL) has the higher concentration of the active ingredient.