Chapter 3 Chemical Compounds
... with different molecular formulas can have the same empirical formulas and such substances will have the same percentage composition. E.g. Acetic acid (C2H4O2), formaldehyde (CH2O), and glucose (C6H12O6) all have the empirical formula CH2O. Generally, empirical formula does not tell us much about a ...
... with different molecular formulas can have the same empirical formulas and such substances will have the same percentage composition. E.g. Acetic acid (C2H4O2), formaldehyde (CH2O), and glucose (C6H12O6) all have the empirical formula CH2O. Generally, empirical formula does not tell us much about a ...
Slide 1
... EXAMPLE: If the empirical formula is CH2O and the molecular mass is 90 g/mole, what is the molecular formula. WE KNOW THE MOLECULAR FORMULA IS SOME WHOLE NUMBER MULTIPLE OF THE EMPIRICAL FORMULA, SO CALCULATE THE EMPIRICAL MASS 1C + 2H + 1O = 12 + 2 + 16 = 30 DIVIDE 90 BY 30 = 90/30 = 3, SO MOLECULA ...
... EXAMPLE: If the empirical formula is CH2O and the molecular mass is 90 g/mole, what is the molecular formula. WE KNOW THE MOLECULAR FORMULA IS SOME WHOLE NUMBER MULTIPLE OF THE EMPIRICAL FORMULA, SO CALCULATE THE EMPIRICAL MASS 1C + 2H + 1O = 12 + 2 + 16 = 30 DIVIDE 90 BY 30 = 90/30 = 3, SO MOLECULA ...
Unit 3 - sotochem
... ➢ Example: Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and acetic acid. If 7.50g of acid and 7.50g of lead (II) acetate are mixed, calculate the number of grams of each reactant and product present after the reaction is completed. ...
... ➢ Example: Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and acetic acid. If 7.50g of acid and 7.50g of lead (II) acetate are mixed, calculate the number of grams of each reactant and product present after the reaction is completed. ...
s - chem116
... Determine the moles of each kind of atom. 88.8 g O / 16 g/mol=5.55 mol O 11.2 g H / 1 g/mol = 11.1 mol H ...
... Determine the moles of each kind of atom. 88.8 g O / 16 g/mol=5.55 mol O 11.2 g H / 1 g/mol = 11.1 mol H ...
Masterton and Hurley Chapter 3
... The compound that gives vinegar its sour taste is acetic acid, which contains the elements carbon, hydrogen, and oxygen. When 5.00g of acetic acid is analyzed it is found to contain 2.00g of carbon, 0.336g of hydrogen, and 2.66g of oxygen. What is the empirical formula of acetic acid? ...
... The compound that gives vinegar its sour taste is acetic acid, which contains the elements carbon, hydrogen, and oxygen. When 5.00g of acetic acid is analyzed it is found to contain 2.00g of carbon, 0.336g of hydrogen, and 2.66g of oxygen. What is the empirical formula of acetic acid? ...
Unit 3 Review Notes - Brinkmann chapter7_and_8_review1
... • Molecules are neutral groups of atoms that are held together by covalent bonds. • Diatomic molecules – H2, N2, O2, F2, Cl2, Br2, and I2. Allotrophs include P4 and S8. ...
... • Molecules are neutral groups of atoms that are held together by covalent bonds. • Diatomic molecules – H2, N2, O2, F2, Cl2, Br2, and I2. Allotrophs include P4 and S8. ...
PPT Empirical & Molecular Formulas
... 1. What is the empirical formula for C4H8? 2. What is the empirical formula for C8H14? 3. Which is a possible molecular formula for CH2O? a) C4H4O4 b) C2H4O2 c) C3H6O3 ...
... 1. What is the empirical formula for C4H8? 2. What is the empirical formula for C8H14? 3. Which is a possible molecular formula for CH2O? a) C4H4O4 b) C2H4O2 c) C3H6O3 ...
Chapter 3: Introduction to chemical formulas and reactivity
... More examples 1. A certain compound is 73.9% Hg and 26.1% Cl by mass. What is the empirical formula? 2. (BLB) A sample of a substance is analyzed and found to contain 5.28 g Sn and 3.37 g F. What is the empirical formula? 3. (BLB) A sample contains 11.66 Fe and 5.01 g oxygen. What is the empirical ...
... More examples 1. A certain compound is 73.9% Hg and 26.1% Cl by mass. What is the empirical formula? 2. (BLB) A sample of a substance is analyzed and found to contain 5.28 g Sn and 3.37 g F. What is the empirical formula? 3. (BLB) A sample contains 11.66 Fe and 5.01 g oxygen. What is the empirical ...
naming-and-formulas-chem-1-ab
... When naming compounds consisting of two nonmetals, the names of the elements are written in the same order as they appear in the formula. (The more metallic element is written first.) ...
... When naming compounds consisting of two nonmetals, the names of the elements are written in the same order as they appear in the formula. (The more metallic element is written first.) ...
Chapter 10b
... point on Earth’s surface at a latitude of 40°N (b) What is the linear speed v of that point? (c) What are w and v for a point on the equator? ...
... point on Earth’s surface at a latitude of 40°N (b) What is the linear speed v of that point? (c) What are w and v for a point on the equator? ...
LACTUER 3 THE MOLECULAR FORMULA / ANALYTICAL
... grams. For example, if 40% of the mass of a compound is oxygen then you calculate you have 40 grams of oxygen. 2. Convert grams to moles. Empirical formula is a comparison of the number of moles of a compound so you need your values in moles. Using the oxygen example again, there are 16.0 grams per ...
... grams. For example, if 40% of the mass of a compound is oxygen then you calculate you have 40 grams of oxygen. 2. Convert grams to moles. Empirical formula is a comparison of the number of moles of a compound so you need your values in moles. Using the oxygen example again, there are 16.0 grams per ...
Unit 9 - Chemical Quantities: The Mole
... Molar Mass • Molar Mass (MM) – Mass of one mole of a substance – Unit is grams/mole • g/mole or g/mol – Same as molecular mass with different unit • molecular mass of water = 18.02 amu • Molar Mass of water = 18.02 g/mol ...
... Molar Mass • Molar Mass (MM) – Mass of one mole of a substance – Unit is grams/mole • g/mole or g/mol – Same as molecular mass with different unit • molecular mass of water = 18.02 amu • Molar Mass of water = 18.02 g/mol ...
4 - Emp. and Mol. Form -q
... Molecular formulas and formula units are always given in whole number values. Divide through by the smallest number of moles in the ratio (in this case that’s 0.5179 mol) to find the whole number ratio of atoms in the empirical formula. Note that showing the mole unit here is optional as all of the ...
... Molecular formulas and formula units are always given in whole number values. Divide through by the smallest number of moles in the ratio (in this case that’s 0.5179 mol) to find the whole number ratio of atoms in the empirical formula. Note that showing the mole unit here is optional as all of the ...
ANGULAR MOMENTUM So far, we have studied simple models in
... with the same value of l, but different values of m. Therefore, the degeneracy is (2l+1). The Spherical Harmonic functions are important in the central force problem--in which a particle moves under a force which is due to a potential energy function that is spherically symmetric, i.e. one that depe ...
... with the same value of l, but different values of m. Therefore, the degeneracy is (2l+1). The Spherical Harmonic functions are important in the central force problem--in which a particle moves under a force which is due to a potential energy function that is spherically symmetric, i.e. one that depe ...
Percent Composition
... REMEMBER THE UNITS FOR MOLAR MASS IS g/mol For example: 1. from moles and grams Find the molar mass if 0.0250 moles of X has a mass of 1.775g ...
... REMEMBER THE UNITS FOR MOLAR MASS IS g/mol For example: 1. from moles and grams Find the molar mass if 0.0250 moles of X has a mass of 1.775g ...
Chapter 3 - Bruder Chemistry
... The quantitative nature of chemical formulas and reactions is called stoichiometry. Lavoisier observed that mass is conserved in a chemical reaction. • This observation is known as the law of conservation of mass. Chemical equations give a description of a chemical reaction. There are two parts to a ...
... The quantitative nature of chemical formulas and reactions is called stoichiometry. Lavoisier observed that mass is conserved in a chemical reaction. • This observation is known as the law of conservation of mass. Chemical equations give a description of a chemical reaction. There are two parts to a ...
CHEMISTRY: HOW MUCH (CALCULATIONS)
... 3. Look at the RATIO to work out how many moles of CuO react with C (which your trying to find) Ratio = 2 (CuO) : 1 (C) Looking at this we can see that there are half the number of moles of C than CuO (2 divided by 2 is 1) so we need to half the number of moles of CuO that there are: 0.5 ÷ 2 = 0.25 ...
... 3. Look at the RATIO to work out how many moles of CuO react with C (which your trying to find) Ratio = 2 (CuO) : 1 (C) Looking at this we can see that there are half the number of moles of C than CuO (2 divided by 2 is 1) so we need to half the number of moles of CuO that there are: 0.5 ÷ 2 = 0.25 ...
Quantum Solutions For A Harmonic Oscillator
... of matrix elements in quantum mechanics. As a follow up, consider the harmonic oscillator problem Hˆ = − ...
... of matrix elements in quantum mechanics. As a follow up, consider the harmonic oscillator problem Hˆ = − ...
Chapter 3
... Stoichiometric relations or ratios may be used to convert between quantities of reactants and products in a reaction. It is important to realize that the stoichiometric ratios are the ideal proportions in which reactants are needed to form products. The number of grams of reactant cannot be directly ...
... Stoichiometric relations or ratios may be used to convert between quantities of reactants and products in a reaction. It is important to realize that the stoichiometric ratios are the ideal proportions in which reactants are needed to form products. The number of grams of reactant cannot be directly ...
Chapter 3. Stoichiometry
... amount of material present in the laboratory (or given in the problem) and the number of moles required by stoichiometry. • Students do not understand that the reagent that gives the smallest amount of product is the limiting reagent. They need much numerical practice at this concept. The use of ana ...
... amount of material present in the laboratory (or given in the problem) and the number of moles required by stoichiometry. • Students do not understand that the reagent that gives the smallest amount of product is the limiting reagent. They need much numerical practice at this concept. The use of ana ...
Percentage Composition
... • Earlier, we found that the empirical formula of a compound was CH2O. • If we know the molecular mass of the compound is 180 g/mol, then we can determine the molecular formula. • We can calculate the mass of the empirical formula (CH2O). It is 30.0 g/mol • Next, we divide our molecular mass of the ...
... • Earlier, we found that the empirical formula of a compound was CH2O. • If we know the molecular mass of the compound is 180 g/mol, then we can determine the molecular formula. • We can calculate the mass of the empirical formula (CH2O). It is 30.0 g/mol • Next, we divide our molecular mass of the ...
Chapter 7: Chemical Formulas and Chemical Compounds
... G. Covalent-Network Compounds 1. Some covalent molecules do not exist of individual molecules. They instead are part of a 3 dimensional network. 2. When this occurs the lowest ratio is given and then named just as binary covalent compounds are. H. Acids and Salts 1. An acid is a distinct type of com ...
... G. Covalent-Network Compounds 1. Some covalent molecules do not exist of individual molecules. They instead are part of a 3 dimensional network. 2. When this occurs the lowest ratio is given and then named just as binary covalent compounds are. H. Acids and Salts 1. An acid is a distinct type of com ...