1. Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: hzhang@tntech.edu

2. CHEM1010/General Chemistry _________________________________________ Chapter 2. (L6)-Introduction to Atoms Today’s Outline..One more property of matter: Density (L3)..Review: Energy, heat, and temperature (L4)..Proust: The law of definite proportions -Is the pure water from Center Hill Lake the same as the pure water from Mississippi River? -The law of definite proportions -Significance of the law of definite proportions

3. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions We all know that there are many many different things. But, now, let’s take a minute asking these questions: Is the pure water from Center Hill Lake the same as the pure water from Mississippi River? Is the table salt produced from the Pacific ocean the same as that from the Atlantic ocean?

4. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions These questions concern if the matter with the same physical properties and chemical properties would be the same universally or not. Perhaps, you think this is so simple, so the answer of course is Yes. But, how do you know? Are you sure? How sure are you?

5. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions Actually, you are right if your answer is Yes. This was verified by many scientists in the early stage of development of modern chemistry. Indeed, many observations by scientists showed that each of many chemicals has the same components and the components are in the same proportions for a particular chemical, regardless of where it comes from or who made it or how it is made.

6. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions This means that a certain chemical always has a fixed proportion of two or more components; in other words, there is a certain mass ratio for the components. Suppose a chemical has two components of A and B, then the mass ratio is fixed, so for example the ratio may be Mass of A : Mass of B = 1:2

7. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions Let’s have a case study: 10.00 g lead + 1.55 g sulfur = 11.55 g lead sulfide 10.00 g lead + 3.00 g sulfur = 11.55 g lead sulfide + 1.45 g sulfur (leftover) 18.00 g lead + 1.55 g sulfur = 11.55 g lead sulfide + 8.00 g lead (leftover) Clearly, the mass ratio is fixed: 10.00 : 1.55 for lead : sulfur

8. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions Let’s have another case study: 32.00 g oxygen gas + 4.00 g hydrogen gas = 36.00 g water 32.00 g oxygen gas + 8.00 g hydrogen gas = 36.00 g water + 4.00 g hydrogen gas 36.00 g oxygen gas + 4.00 g hydrogen gas = 36.00 g water + 4.00 g oxygen gas Clearly, the mass ratio is fixed: 8 : 1 for oxygen gas : hydrogen gas

9. Chapter 2. (L6)-Introduction to Atoms The law of definite proportions Joseph Louis Proust (1754-1826, born in France, living in Spain) did many quantitative analyses about different chemicals and convinced the chemists: In a certain chemical composed of two or more components, the components are always present in a definite proportion by mass. This is the law of definite proportions.

10. Chapter 2. (L6)-Introduction to Atoms Significance of the law of definite proportions This law indicates: -Many chemicals are composed of one or more different components. -The components are in proportion to each other at a fixed ratio in mass, in other words, not random ratio. -The proportion is universal, no matter where and when a chemical is found or produced or in which way it is made, naturally or artificially, by this method or another.

11. Chapter 2. (L6)-Introduction to Atoms Significance of the law of definite proportions The law of definite proportions is a generalization of many many observations by chemists. Any theory to explain why things are different chemically has to be able to account for this law.

12. Chapter 2. (L6)-Introduction to Atoms Quiz Time Given 10.00 g lead + 1.55 g sulfur = 11.55 g lead sulfide, 20.00 g lead + 3.10 g sulfur would result: (a) 11.55 g lead sulfide; (b) 11.55 g lead sulfide + 10.00 g lead; (c) 11.55 g lead sulfide + 1.55 g sulfur; (d) 23.10 g lead sulfide + 0.00 g lead + 0.00 g sulfur

13. Chapter 2. (L6)-Introduction to Atoms Quiz Time Given 32.00 g oxygen gas + 4.00 g hydrogen gas = 36.00 g water, how much of hydrogen gas would be needed to produce 0.18 g water using 0.16 g oxygen gas: (a) 0.2 g; (b) 0.02 g; (c) 0.4 g; (d) 0.1 g

14. Chapter 2. (L6)-Introduction to Atoms Quiz Time We know: oxygen : hydrogen = 8:1, or oxygen = 32 g hydrogen 4 g 32 = 0.16 ; => 8 = 0.16 => 8*X = 0.16* X 4 X X X => 8*X = 0.16 => X = 0.16/8 = 0.02 So, the hydrogen gas needed is 0.02 g. Also, the approach of cutting by half, and then by 1/100.

15. Chapter 2. (L6)-Introduction to Atoms Quiz Time The observation that 10.00 g lead + 1.55 g sulfur = 11.55 lead sulfide is not only an example of the law of definite proportions, but also a good example of (a) the law about emotion; (b) the law about motion; (c) promotion of chemicals; (d) the law of conservation of mass