Lesson
Convert Mass to Moles and Millimoles
Match grams with moles and milligrams with millimoles using the supplied molecular weight.
- Molecular weight
- mol and mmol
- g and mg
- Specified compound form
Specified compound; MW 120 g/mol
1.8 g / 120 g per mol
0.015 mol x 1,000 mmol/mol
Read the molecular weight as a mass relationship
The book expresses the mass of one mole through molecular weight, MW, in g/mol. For a stated MW of 160 g/mol, 160 g represents 1 mol of that specified substance. A millimole is one-thousandth of a mole: 1 mol = 1,000 mmol. The same numerical MW is 160 mg per mmol. Moles and millimoles describe amount of the specified substance; grams and milligrams describe its mass.
Use the mass unit that matches the requested amount
The source formulas are mol = g/MW and mmol = mg/MW, using the supplied numerical MW in g/mol. In an original example, 32 g of a compound with MW 160 gives 32/160 = 0.20 mol. Another example, 480 mg with MW 160, gives 480/160 = 3 mmol. Dividing milligrams by that numerical MW gives mmol, not mol.
Convert the units and verify by a second path
For an original 1.8 g sample with MW 120, convert 1.8 g to 1,800 mg and divide by 120 to obtain 15 mmol. A second path gives 1.8/120 = 0.015 mol, followed by 0.015 x 1,000 = 15 mmol. Keep intermediate precision until the requested final rounding. For 0.72 g with MW 175, 0.0041142857... mol becomes 0.004 mol when three decimal places are requested.
Use the stated form without adding a particle or charge factor
The book examples supply a molecular weight for the compound form being calculated. Use that value; do not replace it with a different form or multiply by dissociation particles when asked for mmol of the compound. The book notes that mEq and mmol are numerically equal for monovalent species. General charge-based mEq calculations are addressed separately; equality must not be assumed for every compound.
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Lesson
Find Millimoles in a Solution
Calculate solute mass in the stated volume before converting it to mmol.
- Percent w/v
- mg/mL
- Total amount
- Amount per volume
4 g A in each 100 mL
Contained in the stated 25 mL
Specified MW 200 g/mol
Start with the solute mass
A % w/v strength gives grams per 100 mL of finished solution. In an original example, 4% w/v Compound A in 25 mL contains 4 g/100 mL x 25 mL = 1 g. With stated MW 200, 1,000 mg/200 = 5 mmol of A. The volume first determines solute mass; it does not replace mass in the MW formula.
Use a mass concentration directly when it is supplied
An original solution contains 12 mg/mL A, MW 120, in 50 mL. Total mass = 12 x 50 = 600 mg; total amount = 600/120 = 5 mmol. The concentration is 5 mmol/50 mL = 0.1 mmol/mL. These are different answers to different requests: 5 mmol is an amount, while 0.1 mmol/mL is an amount per volume.
Scale an amount-per-volume statement
If an original exercise specifies 1.2 mmol of A in each 4 mL, a total 60 mL contains 60 mL x 1.2 mmol/4 mL = 18 mmol. Cancel mL in the proportion. Dividing 1.2 by 4 first gives 0.3 mmol/mL; multiplying that concentration by the total volume gives the same 18 mmol.
Check concentration and total amount separately
For an original 3% w/v solution with MW 150 in final 40 mL, mass is 1.2 g = 1,200 mg and total amount is 8 mmol. Its concentration is 8/40 = 0.2 mmol/mL. Reverse the calculation: 0.2 mmol/mL x 40 mL x 150 mg/mmol = 1,200 mg. Increasing volume at an unchanged strength increases total amount while leaving the concentration unchanged.
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Lesson
Convert Moles or Millimoles Back to Mass
Reverse the MW relationship and scale a per-volume target to the entire preparation.
- Reverse calculation
- mg and g
- Per-volume target
- Stock volume
Specified compound; MW 150 g/mol
90 mL x 0.8 mmol/6 mL
12 mmol x 150 mg/mmol
Multiply amount by molecular weight
Rearrange the book equations: g = mol x MW, and mg = mmol x MW. An original 0.25 mmol target with MW 160 requires 40 mg. An original 0.012 mol target with MW 100 requires 1.2 g. Do not divide the amount by MW when finding mass; division is used in the opposite direction.
Scale the whole preparation before finding mass
An original calculation specifies final 90 mL with 0.8 mmol A per 6 mL and MW 150. The total amount is 90 x 0.8/6 = 12 mmol. Total mass = 12 x 150 = 1,800 mg = 1.8 g. Using only the amount in one 6 mL portion would find the mass for a single portion rather than the complete preparation.
Connect the required mass to a stated stock concentration
For that original 12 mmol target and MW 150, a stock at 30 mg/mL supplies the 1,800 mg in 60 mL. This is the stock volume needed to supply the mass. It does not replace the specified final 90 mL volume. The source calculation establishes the required mass; stock-volume division applies the already-reviewed concentration method to explicitly supplied inputs.
Verify the result with units and the original target
Check 1.8 g = 1,800 mg; 1,800/150 = 12 mmol. Dividing by the final 90 mL gives 0.133333... mmol/mL; multiplying by 6 mL returns the original 0.8 mmol per portion. Keep intermediate precision and round only as instructed. These original exercises teach mass and amount relationships and do not determine a patient dose or establish a complete clinical formulation.
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Module practice
Check the connections.
Each attempt draws 10 questions from the complete 21 question bank.
Each attempt draws a fresh set and rearranges the answer choices.
References
Book source and calculation assumptions.
Original lessons and quantities apply the supplied-book mass-to-mole and reverse-mass methods with explicitly stated molecular weights and solution strengths.
- RxPrep 2023 Course Book
Chapter 1, required-formulas checklist, printed page 7 (PDF page 19); Chapter 10, Moles and Millimoles, printed pages 144-146 (PDF pages 152-154), ending before Milliequivalents.