Unit 1 How do we distinguish substances? The central goal of this unit is to help you understand and apply basic ideas that can be used to distinguish the different substances present in a system. M1. Searching for Differences Identifying differences that allow us to separate components. M2. Modeling Matter Using the particulate model of matter to explain differences. M3. Comparing Masses Characterizing differences in particle’s mass and number. M4. Determining Composition Characterizing differences in particle’s composition.

How do we distinguish substances? Module 3: Comparing Masses Unit 1 How do we distinguish substances? Module 3: Comparing Masses Central goal: To apply the particulate model of matter to determine the relative mass of different particles, as well as the number of particles present in any given sample of a substance.

The Challenge Analysis How much? The particulate model of matter has proven to be a very powerful tool to analyze the properties and behavior of chemical substances. Water How can we use it to differentiate substances in terms of quantitative properties? How can use it to quantify the amount of any substance present in a sample? Carbon dioxide

What is its mass? What is its composition? What is its structure? Focus on Particles If, according to our model, the identity, properties, and behavior of a substance are determined by the nature of its “particles”…. Butane Then, we should focus our efforts on characterizing the properties of these particles. What is its mass? What is its composition? What is its structure?

Indirect Measurements Atoms and molecules have masses and sizes so small that can not be measured directly. The problem has been solved by comparing the masses of samples of different substances with the same number (very large) of particles. H2 2.0 g O2 32.0 g M(O2/H2) = 32.0 g/ 2.0 g = 16. What does this mean?

We can take advantage of our particulate model. Let’s Think However, determining N for every sample is difficult and time consuming. What can we do? We can take advantage of our particulate model. Observe what happens to P as we change the mass of the particles at constant T. How could we use this result to prepare samples of different gases with the same N?

Avogadro’s Law Our results indicate that, at high temperatures and low pressures (where this model is valid): H2 O2

Let’s Think Based on the following information: STP T = 273.15 K P = 1 atm H2 A B2 2.0 g 4.00 g 28.0 g Determine how more massive the unknown atoms A and B are than the H atom.

They contain the same number of particles. Relative Atomic Mass The average relative atomic masses of all known atoms are listed in the periodic table: If you could go to the lab and take 6.94 g of Li, 9.01 g of Be, and 22.99 g of Na, which of the samples will have more or less atoms? Let′s think! “ They contain the same number of particles. But exactly how many?

Avogadro’s Number One mole A sample of any substance with a mass in grams equal to its relative mass has the following number of particles: 26.98 g Al 32.07 g S 24.31 g Mg We call the “molar mass” of a substance the mass of one mole of its particles. No = 6.022 x 1023 particles Avogadro’s Number We call a “mole” the amount of any substance that contains this number of particles.

How would you calculate the molar mass of molecular elements? The molar mass of an atomic element is always equal to its relative atomic mass expressed in grams: Carbon M(Na) = 22.99 g/mol Sodium M(C) = 12.01 g/mol Cl2 Chlorine Phosphorus P4 How would you calculate the molar mass of molecular elements? Let′s think!

In both cases, this is the mass of 6.022 x 1023 molecules Molar Mass The molar mass of a molecular compound is also equal to its relative mass expressed in grams: M(H2O) = 2 x 1.008 + 1 x 16.00 = 18.02 g/mol Water Carbon dioxide M(CO2) = 1 x 12.01 + 2 x 16.00 = 44.01 g/mol In both cases, this is the mass of 6.022 x 1023 molecules

Quantitative Analysis These three concepts are of central importance to answer the question: How much of a substance do we have? C 12.01 amu Molar Mass 12.01 g Atomic Mass Avogadro’s Number Carbon Atom MOLE 6.022 x 1023 units Particulate Macroscopic

Air Properties Let’s apply these ideas to now investigate the properties of our atmosphere in quantitative ways. Density = 1.292 g/L at STP 20.9 O2 78.1 N2 % V Substance Air Composition For example, given this information, how can we calculate the mass in grams of N2 that we can extract from 1 L of air? M(air) = 28.97 g/mol

Air Properties r M First, let’s calculate the moles of air (volume  mass  mole): 4.460x10-2 mol By Avogadro’s hypothesis the %volume is equivalent to the %mole for gases: 78.1% 20.9%

Compare the number of O2 molecules per liter of air at: Let′s think! The following table shows the density of oxygen (O2) at different altitudes. H (km) r (g/L) 0.283 0.653 0.265 1.11 0.254 2.07 0.231 3.65 0.196 4.83 0.173 6.85 0.139 8.25 0.115 12.5 0.069 Compare the number of O2 molecules per liter of air at: Sea level (0 km) Tucson (~0.7 km) Top of Mount Everest (~8.8 km) Air plane cruising (~12 km)

Sea Level To convert from mass to moles, we always need the molar mass of the substances of interest: To convert from moles to number of particles, we always need the Avogadro’s number:

However, their effects are not the same. Air Properties CO2 and CH4 are two well known greenhouse gases. They absorb infrared radiation, trapping thermal energy in the atmosphere. In that sense, they also contribute to global warming. What mass of CH4 is needed to produce the same greenhouse effect as 1.00 g of CO2? Let′s think! However, their effects are not the same. 1 single molecule of CH4 has the same global warming effect as 8 molecules of CO2.

Let′s apply! Assess what you know

Concentrations of O3 above 80 ppb are considered harmful. Let′s apply! Monitoring Ozone Concentrations of O3 above 80 ppb are considered harmful. Ozone (O3), a highly reactive form of oxygen, is a major component of photochemical smog. Its concentration is monitored regularly in major cities. Click on the image to open the website with daily dynamic information about ozone concentrations in Tucson. ppb Parts Per Billion 1 ppb = 1 particle /1x109 particles = 1 L /1x109 L

Let′s apply! Inhaling Ozone On average, people take 24,000 breaths each day.  One single breath has a volume close to 0.5 L. Estimate the number of O3 molecules that go into your body every day. (Use the map to estimate O3 conc.) If we know 1 mol gas ~ 22.4 L at STP

Let′s apply! Estimation Estimate the number of O3 molecules that go into your body every day (25 ppb). Inhaling O3 ozone can result in: Induction of respiratory symptoms (Coughing, throat irritation); Decrements in lung function; Inflammation of airways.

Describe one important application of what you have learned in this module.

For gases at same T and P: Comparing Masses Summary The mass of an atom or molecule is so small that cannot be measured directly. However, we can use the particulate model of matter to determine the relative masses of different particles. For gases at same T and P: Equal V  Equal N H2 Cl2 0.200 g 7.09 g m(Cl/H) = 7.09/0.200 = 35.5

Avogadro’s Number (NA) Comparing Masses Summary We can use relative atomic masses to determine the number o particles in certain mass or vice versa. C 12.01 amu Relative Atomic Mass Molar Mass (M) Avogadro’s Number (NA) MOLE 12.01 g 6.022 x 1023 units Mass (m)   Mol (n)   Number of Particles (N) M NA

How do we know that carbon dioxide is CO2 and not CO3 or C2O4? For next class, Investigate how we could determine the actual composition of the different substances in our atmosphere. How do we know that carbon dioxide is CO2 and not CO3 or C2O4?