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Mass Relationships in Chemical Reactions Atomic Mass (0.9890) (12 amu) + (0.0110) (13.00335) = 12.01 amu.

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Presentation on theme: "Mass Relationships in Chemical Reactions Atomic Mass (0.9890) (12 amu) + (0.0110) (13.00335) = 12.01 amu."— Presentation transcript:

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2 Mass Relationships in Chemical Reactions

3 Atomic Mass (0.9890) (12 amu) + (0.0110) (13.00335) = 12.01 amu

4 Mole Avogadro’s number: 6.022x10 23 (atoms, molecules, particles) Molar mass is numerically the same as atomic mass of an element

5 Molecular mass of a molecule is the sum of all atomic masses x how many atoms Ex) Find the molar mass of H 2 O (1.01 x 2) + 16.00 = 18.02 amu 18.02 amu = 18.02g H 2 O Formula mass of a compound is the sum of all atomic masses x how many atoms Ex) Find the molar mass of NaCl 22.99 + 35.45 = 58.44 amu 58.44 amu = 58.44g NaCl Molar Mass

6 1.Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks and balloons. How many moles of He atoms are in 6.46 g of He? 1.How many atoms are in 112g of Fe? 1.Methane (CH 4 ) is the principal component of natural gas. How many moles of CH 4 are present in 6.07g of CH 4 ? Examples

7 4.How many hydrogen atoms are present in 25.6 g of urea [(NH 2 ) 2 CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers?

8 Mass Spectrometry

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10 Hydrate

11 Percent Composition % comp. = n × molar mass of element molar mass of compound × 100 Ex) Find percent H and percent O in hydrogen peroxide (H 2 O 2 ) %H = 2 × 1.008g H 34.02g H 2 O 2 %O = 2 × 16.00g O 34.02g H 2 O 2 × 100 = 5.926% × 100 = 94.06%

12 Empirical Formula Steps: 1.Percent to mass 2.Mass to moles 3.Divide by small 4.Multiply ‘til whole

13 Example 1. Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92% C, 4.58% H, and 54.50% O by mass. Determine its empirical formula.

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15 Example 2. When ethanol is burned, carbon dioxide and water are given off. Suppose that in one experiment the combustion of 11.5 g of ethanol produced 22.0 g of CO 2 and 13.5 g of H 2 O. Determine the empirical formula for ethanol.

16 Molecular Formula Steps: 1.Find empirical formula 2.Determine the molar mass of the empirical formula 3.Find multiplier (molar mass given/empirical molar mass) 4.Multiply empirical formula subscripts by the multiplier

17 Example A sample of a compound contains 30.46% N and 69.54% O by mass, as determined by a mass spectrometer. In a separate experiment, the molar mass of the compound is found to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound.

18 Chemical Reactions and Equations 2 H 2 (g) + O 2 (g) 2 H 2 O ( l ) coefficient Product(s)Reactants “Reactants with” “to produce” or “yield” State of matter

19 Balancing Chemical Reactions KClO 3 O 2 + KCl Steps: 1.List elements present on each side 2.Add coefficients to balance (“met a non hairy oxen”)

20 Stoichiometry Mole method: coefficient in a reaction can be interpreted as the number of moles Allows us to write conversion factors from a chemical equation

21 Example The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C 6 H 12 O 6 ) to carbon dioxide (CO 2 ) and water (H 2 O): C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O If 856 g of C 6 H 12 O 6 is consumed by a person over a certain period, what is the mass of CO 2 produced?

22 Limiting Reagents CO(g) + 2 H 2 (g) CH 3 OH(g)

23 Example Urea [(NH 2 ) 2 CO] is prepared by reacting ammonia with carbon dioxide in the following process: 2 NH 3 (g) + CO 2 (g) (NH 2 ) 2 CO (aq) + H 2 O ( l ) In one process, 637.2g of NH 3 are treated with 1142g CO 2. Which of the reactants is the limiting reagent? Calculate the mass of [(NH 2 ) 2 CO] formed. How much excess reagent (in grams) is left at the end of the reaction?

24 Reaction Yield

25 Example Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°C and 1150°C: TICl 4 (g) + 2 Mg ( l ) Ti (s) + 2 MgCl 2 ( l ) In a certain industrial operation 3.54 x 10 7 g of TiCl 4 are reacted with 1.13 x 10 7 g of Mg. Calculate the theoretical yield of Ti in grams. Calculate the percent yield if 7.91 x 10 6 g of Ti are actually obtained.


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