General Chemistry Ch. 3 - Stoichiometry of Formulas and Equations
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| Stoichiometry | The study of the quantitative aspects of chemical formulas and reactions
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| Mole (mol) | The SI unit for the amount of substance. It is defined as the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12: 6.022*(10^23) AKA Avogadro’s Number
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| Difference between a mole and a dozen | A mole doesn’t only represent a fixed number of entities, but also a fixed mass. All units of the sample must have the same mass.
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| The central relationship between the mass of one atom and the mass of one mole of those atoms | The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams.
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| Molar Mass (M) | The mass per mole of its entities, thus molar mass has units of grams per mol (g/mol)
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| Molar mass of monatomic elements (E.g. Fe) | Monatomic elements are those that occur as individual atoms. To obtain the molar mass, simply look at the atomic mass. E.g. Fe has a molar mass of 55.85 g/mol
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| Molar mass of molecular elements (e.g. Oxygen) | Molecular elements are those that occur as molecules. To obtain the molar mass of molecular elements, you must scale up the atomic mass by the number of elements in each molecule. E.g. O = 2*(M of O) = 32g/mol
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| Molar mass of compounds (e.g. SO2) | Sum of the molar masses of the elements in the formula. E.g. SO2 = 32.07 + (2*16) = 64.07
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| Isotopic mass | Mass of an isotope of an element
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| When to use Avogadro’s number | To convert between amount (mol) and mass (g), use the molar mass. To convert between amount (mol) and number of entities, use Avogadro’s number (6.022x10^23 entities/mol).
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| Conceptually: How many grams of Ag are in .0342 mol of Ag? | a) How many mol of Ag is in 1g? If one mol is 107.9g, then one g is 1/107.9 mol = .00927mol. b) .0342mol * (1g/.00927mol) = 3.69g Ag.
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| What if the question asks “How many formula units are there in xx.x g of [molecule]? | a) determine molar mass of the molecules in one mol of the solution. b) divide the molar mass of the molecules in the solution by the molar mass of one mol of the molecules to get a ratio. c) mult. ratio by Avogadro’s #.
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| How to calculate Mass % of element using moles | Recall: Mass % = (atoms of X * atomic mass of X) / molecular mass of compound. Using moles: Mass % = (moles of X in formula * molar mass of X) / mass of 1 mol of compound
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| How to take a sample compound expressed in moles and determine the empirical formula of the compound | Take each element’s #moles and divide it by the smallest #moles of the three atoms. Then multiply all three resulting #moles by the smallest number that will turn all #moles into integers.
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| How can we take a sample compound expressed in mass% and figure out both the empirical and molecular formulas? | Change the mass % of each element to gram (e.g. 40% = 40g) thus assuming 100g for simplicity. Convert each to #moles and divide by smallest = empirical. Then (Molar mass sample/Molar mass emp. Formula) and scale up.
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| Combustion analysis | The unknown compound is burned in O2; all of the H is converted to H2O and the C to CO2. By measuring their masses, the third element’s mass can be derived, and the empirical formula can then be derived
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| Can empirical formulas tell us the molecular structure? | Not a lot. Empirical formulas can’t assuredly tell you structure because there are many different compounds with the same empirical formula as they have a scaled-up number of atoms.
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| Can molecular formulas tell us the molecular structure? | No, molecular formulas aren’t able to specify whether a molecule is an isomer, which will affect its structure and function.
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| Isomers | Compounds with the same molecular formula but different properties.
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| Structural (or constitutional) isomer | Atoms are linked together in different arrangements. E.g. Ethanol and Dimethyl Ether; both molecules have the same molecular formula but different structures and different behaviors
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| Chemical equation | A statement in formulas that expresses the identities and quantities of the substances involved in a chemical or physical change.
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| Chemical equations are the “_____” of chemistry; chemical formulas are the “_____”; atomic symbols are the “_____” | “Sentences”; “words”; “letters”
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| The left side of a chemical equation shows _____; the right side of an equation shows _____; for an equation to depict these amounts accurately, it must be _____ | The amount of each substance present before the change; the amount of each substance after the change; balanced, meaning the same number of each type of atom must appear on both sides of the equation
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| 1 mol of H2 gas is mixed with 1 mol of F2 gas and the result is 2 mol of HF. What does this tell us? | H2(g) + F2(g) -> 2HF(g)
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| Reactants | All substances that react during the chemical change, they are placed to the left of the “yield arrow”.
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| Products | All substances produced during the chemical change, they are placed to the right of the “yield arrow”.
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| Balancing (stoichiometric) coefficient: how to insert? | A numerical multiplier of all the atoms in the formula. Start with the most complex substance, place a coefficient 1 in front, then balance by distributing it across the entire equation (similar to debits/credits)
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| After you balance the equations, what should you do? | Adjust the coefficients to turn any fractions into whole numbers by adjusting the entire equation. In most cases, the smallest whole-number coefficients are preferred.
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| After you adjust the coefficients what should you do? | Specify the states of matter.
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| In balancing an equation, although the coefficient specifies the number of molecules/elements, you’re not allowed to alter the… | Chemical formulas (i.e. you can’t alter the subscripts). E.g. If the original molecule is MgO, 2MgO is OK, but you can’t write Mg2O2 even though it has the same number of elements, you’re changing the molecular formula.
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| Combustion reaction | Any compound containing C and H that burns in an excess of air, produces CO2 and H2O
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| If you know the number of moles of one substance, the balanced equation for the reaction tells you _____. Why? | The number of moles of all the others. Why: In a balanced equation, the number of moles of one substance is stoichiometrically equivalent to the number of moles of any other substance
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| Stoichiometrically equivalent | A definite amount of one substance is formed from, produces, or reacts with a definite amount of the other
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| Stoichiometrically equivalent molar ratios | Quantitative relationships between stoichiometrically equivalent substances that we use as conversion factors
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| What can you do if there is a series of chemical reactions that occur in a sequence? | You can create an overall (net) equation. That is cancelling out common terms and adding the series of balanced equations together to arrive at the net equation.
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| Cellular respiration consists of a variety of steps, why is the chemical equation (C6H12O6 + 6O2 -> 6CO2 + 6H2O) relatively simple? | Because the reactants in the series of steps are the products in the following steps. When combined to form a net equation they cancel out.
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| Limiting reactant (or limiting reagent) | The reactant that—when used up—stops the product from forming. Furthermore, it is the reactant that is used up first. Thus the limiting reactant is the one that yields the lower amount of product.
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| Theoretical yield | The amount indicated by the stoichiometrically equivalent molar ratio in the balanced equation. In other words, theoretical yield assumes 100% of the limiting reactant becomes product.
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| How often is theoretical yield actually obtained? | Never
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| Two major reasons theoretical yield ever obtained? | Many reactant mixtures also proceed through one or more side reactions: reactions that form smaller amounts of different products. Also: many reactions seem to stop before they’re complete.
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| Other reasons theoretical yield is never obtained | Small amounts of product binding to containers, distillate evaporating, etc. Laboratory tools can rarely get every bit of reactant to react.
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| Actual yield | The amount of product that you actually obtain
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| Percent yield (% yield) | The actual yield expressed as a percentage of the theoretical yield: (actual/theoretical * 100)
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| Calculating the percent yield in multistep syntheses | Each step is expressed as a fraction of 1.00 and multiplied by the others then 100 to get the percent yield. E.g. for a six-step reaction that has a 90% yield each step, the overall recovery will only be 50%.
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| Green chemistry | A field that is focused on developing methods to synthesize compounds efficiently and reduce or prevent the release of harmful products into the environment
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| Atom economy | The proportion of reactant atoms that end up in the desired product. It is a useful criterion for choosing the most efficient synthetic route
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| % atom economy | (# moles * molar mass of desired product) / (sum of #moles*molar mass for all products)
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| Solute | The substance dissolved in a larger amount of another solution
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| Solvent | The larger amount of solution in which the solute is dissolved
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| Solution | The homogeneous mixture containing the solute and solvent. The solute’s individual chemical entities become evenly dispersed throughout the available volume and surrounded by solvent molecules
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| Concentration | The amount of solute dissolved in a given amount of solution. It is an intensive quantity (such as density of temperature) and thus independent of the volume of the solution.
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| Molarity (M) | Expresses the concentration in units of moles of solute per liter of solution: (mol solute)/(L soln)
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| Molarity can be thought of as | A conversion factor used to convert between volume of solution and amount of solute, from which we then find the mass or the number of entities of solute.
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| Remember that the volume term in the denominator of molarity is the _____ volume, not the _____ volume. | Solution; solvent. Why: the solution includes solute + solvent. Solvent doesn’t include solute.
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| How to go from higher molarity to lower molarity | A concentrated solution (higher molarity) is converted to a dilute solution (lower molarity) by adding solvent to it.
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| Isotonic saline | A .15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids.
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| Formula which can be manipulated to solve dilution problems, and others involving a change in concentration | M_dil * V_dil = #moles = M_conc * V_conc
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| How do antacids word | Specialized cells in the stomach release HCl to aid digestion. Too much HCl released can be neutralized with magnesium hydroxide Mg(OH)2, which reacts with the acid to form water and magnesium chloride MgCl2
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