1 / 42

Regents Chemistry Stoichiometry

Regents Chemistry Stoichiometry. 02. Atomic Mass Unit (amu). amu- unit used to for mass of an atom amu of Oxygen? 16amu Why not in grams? Mass of oxygen is really 2.7 × 10 -23 g Atoms are too small, number is too bulky. Find the mass of the following atoms. Mg = ______ Li = ______

vila
Télécharger la présentation

Regents Chemistry Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Regents Chemistry Stoichiometry

  2. 02 Atomic Mass Unit (amu) • amu- unit used to for mass of an atom • amu of Oxygen? 16amu • Why not in grams? • Mass of oxygen is really 2.7 × 10-23g • Atoms are too small, number is too bulky

  3. Find the mass of the following atoms • Mg = ______ • Li = ______ • Cl = ______ 4. Al = ______ 5. Ca = ______ 6. H = ______ 24 amu 23 amu 7 amu 20 amu 35 amu 1 amu

  4. Types of Elements Monoatomic Element • One atom of an element that is stable enough to stand on its own (VERY RARE) • not bonded to anything Diatomic Elements • Elements whose atoms always travel in pairs (Br,I,N,Cl,H,O,F) • Bonded to another atom of the same element.

  5. Formula Mass • mass of an atom, molecule, compound • O2 • This means that mass of O2 = 2 × _______ amu = ________ amu • Notice how we use rounded numbers Subscript = tells you the total number of atoms in the compound/molecule. 16 32

  6. Gram Formula Mass

  7. Formulas • Molecular – actual ratios of atoms in a molecule or compound. • ex: C4H10 or C6H12O6 • Empirical – simplest ratio of atoms in a compound • ex: C2H5 or CH2O • Structural – shows the arrangement of the atoms in the actual compound

  8. Example • The empirical formula is CH and the molecular mass is 26. What is the molecular formula? • C2H2 • C3H3 • C4H4 Step 1: find mass of empirical CH = 13amu Step 2: Divide molecular mass by empirical mass 26 (given)/13 = 2 Step 3: multiply empirical formula by answer in step 2 CH × 2 = C2H2

  9. Review - The molecular formula is C3H6. What is the empirical formula? ______________ - Which is an empirical formula? • C2H2 • H2O • H2O2 • C6H12O6 • What is the empirical formula of the compound whose molecular formula is P4O10? • PO • PO2 • P2O • P8O20

  10. Chemical Equation • Shows which bonds are broken and which bonds are built. • Numbers of atoms on the left side must equal number of atoms on the right side of the arrow • After the elements are correctly written, only the coefficient can be changed. • No coefficient means there is only one molecule H2 + O2 H20 2 atoms H 2 atoms O 2 hydrogen atoms 1 oxygen atom 2 2 4 = 4 = 2 =

  11. Fix this equation! • Formationof salt from sodium and chlorine gas. Na + Cl2 → NaCl Na Cl Na Cl Cl PRODUCES

  12. Modeling Conservation of Matter

  13. Review equations Mg + Cl2 MgCl2 Ca + HCl  CaCl2 + H2 Ca + H20  Ca(OH)2 + H2 Given the incomplete equation: 2N2O5(g)  Complete the balanced equation. • 2N2(g) + 3H2(g) • 2N2(g) + 2O2(g) • 4NO2(g) + O2(g) • 4NO(g) + 5O2(g)

  14. Reaction Types Four Basic Types • Synthesis • Decomposition • Single replacement (substitution) • Double replacement

  15. Synthesis • Formation of only ONE product from two reactants, but not always. Examples: N2(g) + 3H2(g) 2NH3(g) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) because O2 combines with both metal and nonmetal to form two oxides.

  16. Decomposition • One reactant breaks apart to form several products. • AKA combustion if products are CO2 and H2O Examples: 2H2O2(aq) 2H2O(l) + O2(g) hydrogen peroxide decomposes over time to leave behind water.

  17. Single Replacement • A more active metal replaces a less active metal in a compound, or vice versa. Example 2Fe(s) + 6HCl(aq)  2FeCl3(aq) + 3H2(s) what happens when a metal becomes corroded by an acid where an element is reacting with a compound.

  18. Double replacement • Reaction between aqueous compounds. • Cations and anions switch position. • If an insoluble precipitate forms, the reaction is an end reaction, otherwise an aqueous mixture of ions Example AgNO3 (aq) + NaCl(aq)  NaNO3(aq) + AgCl(s)

  19. Review • Cl2(g) + 2NaBr(aq)  2NaCl(aq) + Br2(l) 2. FeCl3(aq) + 3NaOH(aq)  Fe(OH)3(s) + 3NaCl(aq) 3. 2Mg(s) + O2(g)  2MgO(s) 4. H2CO3(aq)  H2O(l) + CO3(g) SR DR S D/Combustion

  20. Gram Formula Mass (GFM) AKA • Molar Mass • Atomic mass • Molecular mass (only in covalent) • Mass of formula in grams

  21. Formula Mass • Sum of atomic masses in the molecule • What is the formula mass (or molecular mass) of K2CO3?

  22. gram formula mass (GFM) • GFM – describes the mass of one mole of a compound • To find GFM, add individual GAM for each element in the compound • H20 = 1.00 + 1.00 + 16.00 = 18.00 g • (NH4)2SO4 = N = 2 × 14 = 28 H = 8 × 1 = 8 S = 1 × 32 = 32 O = 4 × 16 = 64 132 g

  23. Gram formula Mass (molar mass)= mass in grams • Mass of 6.02 x 1023 particles (1 mole of particles). • If you weigh 6.02 x 1023 particles (1 mole of particles) of K2CO3, the weight on the scale will be 138 grams

  24. Molecular Mass H20 = 1.01 + 1.01 + 16.00 = 18.02 g/mol (NH4)2SO4 = N = 2 × 14.00 = 28 H = 8 × 1.01 = 8.08 S = 1 × 32.00 = 32 O = 4 × 16.00 = 64 132.08 g/mol

  25. Determine the molecular Formula of the following HNO3 H N O 1×1 1×14 3×16 1 + 14 + 48 63amu (NH4)2CO3 N H C O 2×14 8×1 1×12 3×16 28 + 8 + 12 + 48 96amu

  26. % Composition HNO3 63amu H N O 1/63 14/63 48/63 ×100 ×100 ×100 1.59% 22.22% 76.19% Needs to total 100% (NH4)2CO3 96amu N H C O 28/96 8/96 12/96 48/96 ×100 ×100 ×100 ×100 29.17% 8.33% 12.50% 50.00% Needs to total 100%

  27. % Given instead • Cmpd is 86% C and 14% H. What is the empirical formula? C H 86/12 14/1 7.17 C 14 H 7.17/7.17 14/7.17 1 2 CH2 % of water in Na2CO3 • 10H2O (formula mass = 286)? H2O = 2 + 16 = 18amu 10 × 18am = 180 180/286 × 100 = 62.94%

  28. Rule 1 mole (of molecules) Equals 1 gram molecular mass Equals 6.02 x 1023 molecules Equals 22.4 liters (for gases at STP)

  29. Example • What is the formula mass of NO2? 46 gram • What is the mass of 2 moles of NO2? 2 X 46 gram = 92 grams • What is the mass of 12 x 1023 molecules of NO2? 12 x 1023 / 6.02 x 1023 = 2 2 x 46 = 92 grams • What is the mass of 44.8 liters of NO2? 44.8 / 22.4 = 2 2 x 46 = 92 grams

  30. Density Density = mass / volume Usually expressed for gases in grams/liter Gram formula mass = density at STP (g/L) x 22.4 liters A piece of aluminum has a mass of grams and a volume of ml. What is its density? 1.35 g/l x 22.4 l = 30.24 grams

  31. Another example Which gas has a density of 1.70g/l at STP? • F2 • He • N2 • SO2 Total the mass of each choice to find the answer.

  32. Mole-Mole Problems • Answers how many moles of one element or compound react with a given number of moles of another element or compound. • How many moles of Ca are needed to react completely with 6 moles of H2O in the following reaction: Step 1: Balance equation Step 2: Cross out molecules not needed. Step 3: Write mole number on top of given formula and an x on the unknown Step 4: Write mole number on bottom of formula from balanced equation Step 5: set up proportions x 6 Ca + H2O  Ca(OH)2 + H2 2 1 mole 2 mole

  33. Review/Practice • Given the reaction: CH4 + O2 CO2 + H2O • How many moles of oxygen are needed for the complete combustion of 3.0 moles of CH4? • 6.0 moles • 2.0 moles • 3.0 moles • 4.0 moles • What amount of oxygen is needed to completely react with 1 mole of CH4? • 2 moles • 2 atoms • 2 grams • 2 molecules

  34. Avogadro’s number • measured to be approximately 6.022 x 1023 (to 4 s.f) • Chemists use the mole in the same way that grocers use the dozen for groups of 12 and stationers use the ream for groups of 500.  •  we can use the mole without being overly concerned about exactly how many objects it represents

  35. Mole • smallest measurable mass of matter contains trillions of atoms, so chemists use a unit of amount called the mole (abbreviated mol). • one mole is the number of atoms in 12 g of carbon-12 • one atom of tin-120 has a mass of 120 u, it follows that one mole of tin-120 atoms will have ten times the mass of one mole of carbon-12 atoms, i.e. 120 g.  • the mass in grams of one mole of atoms of any element will be numerically equivalent to its atomic mass in g/mol.

  36. DIMO

  37. Dimensional Analysis

  38. Review • Which gas sample contains a total of 3.0 x 1023 molecules? • 71g of Cl2 • 2.0 g of H2 • 14g of N2 • 38g of F2 A sample of an unknown gas at STP has a density of 0.630 g/l. What is the gram molecular mass of this gas? • 2.81g • 14.1g • 22.4g • 63.0g • Which quantity represents 0.500 mole at STP? • 22.4L of Ar • 11.2L of N2 • 32.0L of H2 • 44.8L of He

  39. Using Avogadro’s number • how many atoms are in a sample of silicon that has a mass of 5.23 g.  • Moles can be number of atoms or particles in a molecule

  40. Mole • mass of 1 atom = mass of a mole of atoms / 6.022 x 1023 • mass of 1 C atom = 12.01 g / 6.022 x 1023 C atoms • mass of 1 C atom = 1.994 x 10-23 g

  41. 04

More Related