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- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/00%3A_Front_Matter
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.4%3A_Percentage_YieldStoichiometric calculations will give you a theoretical yield for a reaction; the yield that you should obtain assuming that the reaction proceeds with 100% efficiency and that no material is lost in ...Stoichiometric calculations will give you a theoretical yield for a reaction; the yield that you should obtain assuming that the reaction proceeds with 100% efficiency and that no material is lost in handling. The amount of material that you isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The percentage of the theoretical yield that you actually isolate is called the percentage yield.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/11%3A_Nuclear_Chemistry/11.3%3A_Beta_Particle_EmissionAdding the atomic numbers on the right side of the equation shown above gives {(-1) + (+1) = 0}; identical to the “atomic number” in the neutron (n10); (even though a neutron can break do...Adding the atomic numbers on the right side of the equation shown above gives {(-1) + (+1) = 0}; identical to the “atomic number” in the neutron (n10); (even though a neutron can break down to produce a proton, there are no actual protons in a neutron, hence its atomic number is zero). Because the neutron is converted into a proton, the atomic number of the element increases by one unit, changing the identity of the element to the next highest in the periodic table.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/04%3A_The_Mole_and_Measurement_in_Chemistry/4.5%3A_Empirical_and_Molecular_FormulasMany of the methods, however, that chemists use in the laboratory to determine the composition of compounds do not give the molecular formula of the compound directly, but instead simply yield the low...Many of the methods, however, that chemists use in the laboratory to determine the composition of compounds do not give the molecular formula of the compound directly, but instead simply yield the lowest whole-number ratio of the elements in the compound. For example, the molecular formula for glucose is C 6 H 12 O 6 , but the simplest whole-number ratio of the elements in glucose is CH 2 O; if you multiply each element in (CH 2 O) by six, you obtain the molecular formula for glucose.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/10%3A_Principles_of_Chemical_Equilibrium/10.3%3A_Calculating_Equilibrium_ValuesIn the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to domi...In the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to dominate the equilibrium mixture, making the ratio
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/03%3A_Chemical_Bonding_and_Nomenclature/3.S%3A_Chemical_Bonding_and_Nomenclature_(Summary)When naming simple, binary ionic compounds, the cation is named first using the name of the element, followed by the anion, where the suffix ide is added to the root name of the element. For transitio...When naming simple, binary ionic compounds, the cation is named first using the name of the element, followed by the anion, where the suffix ide is added to the root name of the element. For transition metals in which the metal can assume a variety of oxidation states (different positive charges), the charge of the metal ion is shown in the name using Roman numerals, in parenthesis, following the name of the element (i.e., iron (III) chloride).
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/05%3A_Chemical_Reactions/5.S%3A_Chemical_Reactions_(Summary)In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (t...In a chemical equation, the chemical formulas for the substance or substances that undergo the chemical reaction (the reactants) and the formulas for the new substance or substances that are formed (the products) are both shown, and are linked by an arrow. The arrow in a chemical equation has the properties of an “equals sign” in mathematics, and because of this, in a chemical equation, there must be the same number and types of atoms on each side of the arrow.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_ChemistryWhile the notion of chemistry and math (together in the same room) may make you want to scream, we will see in this chapter that concepts such as chemical stoichiometry and mass balance are not overwh...While the notion of chemistry and math (together in the same room) may make you want to scream, we will see in this chapter that concepts such as chemical stoichiometry and mass balance are not overwhelming and can be approached using the same problem-solving algorithms that we have mastered in previous chapters. Limiting reactant problems may appear challenging, but we will see it is the same calculation that we do routinely… we simply have to do the calculations twice.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.2%3A_Molar_Stoichiometry_in_Chemical_EquationsIn terms of mass, two moles of sodium, having a total mass of 45.98 grams, would react with one mole of chlorine gas (a mass 70.90 grams) to give two moles of sodium chloride, for a total of 275.9 gra...In terms of mass, two moles of sodium, having a total mass of 45.98 grams, would react with one mole of chlorine gas (a mass 70.90 grams) to give two moles of sodium chloride, for a total of 275.9 grams of product. As an example, return to the question of the decomposition of H 2 O 2 . If 0.28 moles of H 2 O 2 decompose, according to the equation given below, how many moles of oxygen gas (O 2 ) will be formed?
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/06%3A_Quantitative_Relationships_in_Chemistry/6.5%3A_Limiting_ReactantsYou may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is ...You may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is determined by the limiting reactant in the reaction, and some of the excess reactant is left over.
- https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/02%3A_The_Physical_and_Chemical_Properties_of_Matter/2.5%3A_Conservation_of_MassThe law of mass conservation states that there is no detectable change in the total mass of materials when they react chemically to form new materials.
