(1.864 g) (137.33 g/mol / 233.39 g/mol) = 1.0968 gĢ) Calculate mass of anhydrous BaCl 2 that contains 1.0968 g of Ba:ġ.0968 g is to 137.33 g/mol as x is to 208.236 g/molģ) Calculate mass of water in original sample:Ĥ) Calculate moles of anhydrous BaCl 2 and water:ĥ) Express the above ratio in small whole numbers with BaCl 2 set to a value of one:Įxample #5: Given that the molar mass of Na 2SO 4 nH 2O yields 1.864 g of anhydrous BaSO 4 after treatment with sulfuric acid, calculate n.This means that the HCl reacted with 0.013722 mole of sodium carbonate.Ġ.013722 mol times 105.988 g/mol = 1.4544 gĢ.4723 g / 18.015 g/mol = 0.13724 mol of waterĦ) For every one Na 2CO 3, how many waters are there?Ĭomment: this is one of the three sodium carbonate hydrates that exist. We also assume that no CO 2 dissolves in the water.)Ġ.6039 g / 44.009 g/mol = 0.013722 mol of CO 2ģ) Use the 1:1 molar ratio referenced above: (Also, note that we assume that the gas is pure CO 2 and that there is no water vapor whatsoever. The key is that there is a 1:1 molar ratio between Na 2CO 3 and CO 2. Ignore the water of hydration for a moment. What is the number of water molecules bonded to Na 2CO 3 (value of n)? nH 2O with excess HCl(aq), 0.6039 grams of a gas is given off.Just sayin'.Įxample #3: When you react 3.9267 grams of Na 2CO 3 This is a problem probably crafted so that you cannot look up possible answers via the InterTubez®. Na 2CO 3 -> 0.0304 mol / 0.0304 mol = 1Ĭomment: sodium carbonate forms three hydrates and the above is not one of them. Determine the formula of the hydrate and then write out the name of the hydrate.Ģ) Determine moles of Na 2CO 3 and water: After heating, the mass of the anhydrous compound is found to be 3.22 g. H 2O -> 8.09 g / 18.015 g/mol = 0.449 molĮxample #2: A hydrate of Na 2CO 3 has a mass of 4.31 g before heating. The original location of this document is here.Įxample #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. Here's a worksheet with eight hydrate problems, all of which have (hand-written) solutions. Ten problems Calculate empirical formula when given mass data Fifteen problems Calculate empirical formula when given percent composition data A list of all the problems Determine identity of an element from a binary formula and a percent composition Mole Table of Contents Determine identity of an element from a binary formula and mass data The difference in the mass of the hydrate and the mass of the anhydrous salt can be used to determine the formula of the compound.ChemTeam: Determine the formula of a hydrate: fifteen examplesĭetermine the formula of a hydrate: fifteen examples The mass of the resulting anhydrous salt is determined. The premise of the lab is simple: A known mass of a hydrate is heated to release the water of hydration. So this past week we did a lab called " Composition of Hydrates." Hydrates are compounds that have some number of water molecules attached to them. Lately, I have been teaching how to write chemical formulas and the naming of compounds. and I am being reminded how much I love them! I love the mathematical and analytical nature of chemistry labs and the need for exact and precise laboratory procedures. So this year I am doing chemistry labs that I have not done for some time. My school made the transition this year from a 5 period day to a 7 period day, thus the need to assign teachers additional classes to teach. My usual teaching assignment is a full day of AP Biology. This is the first time in about 7 years that I have taught a chemistry class.
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