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% COMPOSITION BY MASS

“Women and cats will do as they please, and men and dogs should relax and get used to the idea.” ― Robert A. Heinlein. SOLUTIONS CONSIST OF TWO PARTS – THE SOLVENT (THE STUFF DOING THE DISSOLVING) AND THE SOLUTE (THE STUFF BEING DISSOLVED).

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% COMPOSITION BY MASS

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  1. “Women and cats will do as they please, and men and dogs should relax and get used to the idea.” ― Robert A. Heinlein

  2. SOLUTIONS CONSIST OF TWO PARTS – THE SOLVENT (THE STUFF DOING THE DISSOLVING) AND THE SOLUTE (THE STUFF BEING DISSOLVED). USUALLY, WE WANT TO KEEP TRACK OF HOW MUCH SOLUTE IS CONTAINED IN A GIVEN AMOUNT OF SOLUTION. WE REFER TO THIS AS THE SOLUTION CONCENTRATION. THERE ARE SEVERAL DIFFERENT WAYS OF EXPRESSING THIS.

  3. % COMPOSITION BY MASS THIS IS THE MASS OF THE SOLUTE DIVIDED BY THE MASS OF THE SOLUTION (SOLUTE PLUS SOLVENT) DIVIDED BY 100. FOR EXAMPLE: IF YOU HAVE 20 g OF NaCl IN 100 g OF SALT SOLUTION. % COMPOSITION = (20 g / 100 g) x 100 % COMPOSITION BY MASS = 20 % % BY MASS IS USUALLY JUST EXPRESSED AS %.

  4. % COMPOSITION BY VOLUME OR % v/v THIS IS USED SOMETIMES WHEN DEALING WITH SOLUTIONS OF LIQUIDS DISSOLVED IN LIQUIDS. THIS IS DETERMINED BY DIVIDING THE VOLUME OF THE SOLUTE BY THE VOLUME OF THE SOLUTION AND MULTIPLYING BY 100. THE VOLUME OF THE SOLUTION IS NOT NECESSARILY THE VOLUME OF THE SOLVENT PLUS THE VOLUME OF THE SOLUTE.

  5. WINE IS ABOUT 12% v/v ETHANOL. THIS MEANS THAT FOR EVERY 100 ml OF WINE, THERE ARE 12 ml OF ETHANOL. IF YOU MIX 12 ml OF ETHANOL AND 88 ml OF WATER, YOU WILL GET LESS THAN 100 ml OF WINE. VOLUMES OF LIQUIDS ARE NOT ADDITIVE.

  6. MOLE FRACTION (X) THE MOLE FRACTION OF A COMPONENT IS THE NUMBER OF MOLES OF THAT COMPONENT DIVIDED BY THE TOTAL NUMBER OF MOLES OF ALL THE COMPONENTS IN THE SOLUTION. NOTE THAT THE MOLE FRACTIONS OF ALL THE COMPONENTS IN A SOLUTION SHOULD ADD UP TO 1.

  7. EXAMPLE: 22 g OF NaCl and 15 g OF KCL ARE DISSOLVED IN 200 ml OF WATER. WHAT ARE THE MOLE FRACTIONS OF ALL COMPONENTS

  8. ONE OF THE MORE COMMON UNITS FOR EXPRESSING SOLUTION CONCENTRATION IS MOLARITY. MOLARITY IS DEFINED AS THE NUMBER OF MOLES OF A SOLUTE PER LITER OF SOLUTION. M = # moles solute/liters of solution

  9. EXAMPLE: What is the molar concentration of NaCl in a solution prepared by dissolving 35.2 g NaCl in enough water to produce 500 ml of solution? Mass of NaCl Molar mass of NaCl Moles of NaCl Liters of solution molar mass NaCl = 58.44 g/mole moles NaCl = (35.2 g)/(58.44 g/mole) = 0.602 moles volume NaCl solution = 500 mL = 500 mL x (1 L/1000 mL) = 0.5 L solution Molarity NaCl = (0.602 moles/0.5 L) = 1.20 moles NaCl/L solution = 1.20 M Molarity of NaCl solution

  10. This is the equipment that you would normally use to prepare a solution of a given molar concentration. • First, you would accurately weigh the solute • Then, you would transfer the solute to the volumetric flask. You might have partially filled the volumetric flask with solvent. • Next, you would stir the mixture until the solvent had dissolved. • Finally, you would carefully fill the volumetric flask with solvent to the calibration line and mix the solution.

  11. http://www.uwplatt.edu/chemep/chem/chemscape/labdocs/catofp/mixpour/volflask/volflask.htm#descriptionhttp://www.uwplatt.edu/chemep/chem/chemscape/labdocs/catofp/mixpour/volflask/volflask.htm#description AT YOUR CONVENIENCE, YOU SHOULD GO TO THIS SITE AND READ THROUGH WHAT IS INVOLVED IN SOLUTION PREPARATION. DON’T FORGET TO DO THE FINAL MIX ON THE SOLUTION BY INVERTING THE FLASK SEVERAL TIMES.

  12. The equipment is usually accurately calibrated enough so that your measurements will be accurate to four significant figures.

  13. EXAMPLE: What is the molar concentration of KBr in a solution prepared by dissolving 0.321 g of KBr in enough water to form 0.250 L solution?

  14. OFTEN IN THE LAB YOU WILL FIND YOURSELF DILUTING A MORE CONCENTRATED SOLUTION TO PREPARE A LESS CONCENTRATED SOLUTION. FOR EXAMPLE, MOST ACIDS ARE DELIVERED IN CONCENTRATED FORM. STOCK HYDROCHLORIC ACID, HCl = 12 M STOCK ACETIC ACID, CH3COOH = 17.1 M STOCK SULFURIC ACID, H2SO4 = 18 M STOCK AMMONIUM HYDROXIDE, NH4OH = 14.5 M THESE VALUES WILL VARY SOME, SO IT IS A GOOD IDEA TO CHECK THE LABEL, BUT THEY ARE GOOD STARTING ESTIMATES.

  15. IN A DILUTION, YOU ARE STARTING WITH A CERTAIN VOLUME AND CONCENTRATION OF SOLUTION, AND YOU ARE ADDING SOLVENT TO MAKE IT MORE DILUTED. NOTE: YOU ARE NOT CHANGING THE AMOUNT OF SOLUTE. MOLARITY = #moles/#liters = m/V IN A DILUTION, YOU ARE ONLY CHANGING V. MOLARITY X VOLUME = #moles = M X V SO, IN A DILUTION #moles1 = #moles2 OR M1 x V1 = M2 x V2

  16. EXAMPLE: YOU WANT TO PREPARE 250 ml OF 1 M HCl. HOW MANY ml OF CONCENTRATED HCl (12 M) DO YOU NEED TO DILUTE TO 250 ml? M1V1 = M2V2 M1 = 12 M V1 = unknown M2 = 1 M V2 = 250 ml SO, M1V1 = M2V2 (12 M) X V1 = (1 M) X 250 ml V1 = 250 / 12 = 20.8 ml NOTE: AS LONG AS YOU USE THE SAME UNITS ON BOTH SIDES OF THE EQUATION, YOU DO NOT NEED TO CONVERT.

  17. Molality, m – the number of moles of solute per kilogram of solvent molality = quantity of solute (mol) / kilograms of solvent

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