Worked example
A 0.300 g pure analyte portion requiring 50.0 mL of 0.100 mol/L titrant, with a 1:1 reaction ratio, represents 0.00500 mol analyte. The apparent molar mass is 60.0 g/mol.
The balanced reaction supplies the ratio
OpenStax’s quantitative-analysis chapter calculates analyte amount from the known stoichiometry between analyte and titrant. Enter coefficients for those two species from the same balanced equation. A diprotic acid, for example, is not automatically 1:1 with a monobasic titrant; the actual reaction must establish the ratio.
Mass and titre must represent the same portion
Enter the analyte mass represented by the titrated portion, not an entire sample mass when only an aliquot was titrated. The calculator does not adjust for dilution, aliquot fraction, blanks, hydration or impurities. Correct those quantities using your experiment’s documented procedure before entering them.
An apparent result is not an identification
The result assumes complete reaction of the specified analyte, valid equivalence-volume measurement and the stated concentration. An experimental endpoint may differ from the equivalence point. Impurities or the wrong ratio can change the apparent mass substantially; the number alone does not identify the substance.
Explore the related calculation
For a different starting point, try the molarity-to-moles calculator or return to the main calculator. Browse the complete tool directory for this site.
Reference and corrections
The definitions and assumptions above follow these sources; the examples and calculator arithmetic are original to this page.
Found a problem? Email contact@gramstomoles.org.