1. Conservation of Mass 𝑚 𝑖𝑛 𝑚 𝑜𝑢𝑡
𝑚 𝑖𝑛 − 𝑚 𝑜𝑢𝑡 + 𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 − 𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 = ∆ 𝑚 𝑠𝑦𝑠𝑡𝑒𝑚

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3. Material Balance
Engineers design products, and we make them out of different raw materials.
change the shape of a body by adding or removing material
change the organization of atoms in a material
combine components to create new substances through chemical reactions
removing material
Blacksmith hot forging a steel tool using a hammer and an anvil.
rearranging atomic structure
OH-
H2O H e-
2 𝐻 2 𝑂(𝑙)+2 𝑒 − → 𝐻 2 𝑔 +2 𝑂𝐻 − (𝑎𝑞)
chemical reactions
We use material balance to analyze all sorts of physical processes (industrial, biological, environmental)
mass can neither be created nor destroyed just rearranged
short of nuclear reactions (𝐸=𝑚 𝑐 2 )

4. Material Balance - Keeping Track of the Mass
2 𝐻 2 𝑂(𝑙)+2 𝑒 − → 𝐻 2 𝑔 +2 𝑂𝐻 − (𝑎𝑞)
If chemical reactions occur, new system components may be generated
𝑚 𝑖𝑛 − 𝑚 𝑜𝑢𝑡 + 𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 − 𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 = ∆ 𝑚 𝑠𝑦𝑠𝑡𝑒𝑚
accumulation of mass in the system
massfinal – massinitial
While others are consumed
𝑚 𝑖𝑛
Example Applications ( 𝑚 𝑖𝑛 − 𝑚 𝑜𝑢𝑡 =∆ 𝑚 𝑠𝑦𝑠𝑡𝑒𝑚 )
water in a lake
water-flowingin + rain – water-flowingout - evaporation ≈∆ water mass
laundry dryer
wet-laundryin – partially-dry-laundryout - evaporation = 0
fishtank project
NaCl initially in system + NaCl added - NaCl leaving through overflow = ∆ NaCl
𝑚 𝑜𝑢𝑡

5. Batch Problems Start with nothing and end with nothing in system
making a batch of homemade ice cream
mixing a batch of concrete in a mixer
Example Problem: A 10-gallon aquarium contains 2% salt by weight. How much salt would you need to add to bring the salt concentration to 3.5% salt by weight?
1. Draw a diagram to represent the system
3. Convert volumes to mass
2. Label all inputs and outputs, assigning variables to unknowns
𝑋 lbs dry salt
𝑊 𝐻 2 𝑂 =10𝑔𝑎𝑙
∙ 𝑓 𝑡 3 𝑔𝑎𝑙
∙62.3 𝑙𝑏 𝑓 𝑡 3
=83.3 𝑙𝑏
10 gal salt water
2% NaCl
98% H2O
𝑌 lbs salt water
3.5% NaCl
96.5% H2O

6. Example Problem continued
𝑋 lbs dry salt
𝑚=83.3 𝑙𝑏
10 gal salt water
2% NaCl
98% H2O
𝑌 lbs salt water
3.5% NaCl
96.5% H2O
4. Apply conservation of mass to each component (salt & water) and for mixture:
𝑚 𝑖𝑛 − 𝑚 𝑜𝑢𝑡 + 𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 − 𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 = 𝑚 𝑓𝑖𝑛𝑎𝑙 − 𝑚 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
overall:
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
salt:
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
water:
83.3 𝑙𝑏
+ 𝑋 =𝑌
0.02(83.3 𝑙𝑏)
+ 𝑋
=0.035∙𝑌 𝑙𝑏
=0.965∙𝑌
(1)
(2)
(3)
4. Solve for the unknowns (you can use any of the three equations above)
From (3):
𝑌= 0.98 (83.3 𝑙𝑏) =84.6 𝑙𝑏
Is equation (2) useful??? why??
Plug this into (1):
𝑋=𝑌−83.3 𝑙𝑏
=84.6 𝑙𝑏 −83.3 𝑙𝑏=1.29 𝑙𝑏
use it to check your work
dry salt to add

7. Problem Solving Tips
Draw a picture of the system. Sometimes it’s not easy to determine the boundaries of your system. (a large river flowing into the ocean for example where does river end and ocean begin?)
Label all inputs and outputs, listing all known quantities & concentrations and assigning variables to the unknowns. This key step is where errors usually occur.
Think about the problem a little bit. What is happening with the overall system? Are components generated or consumed? Revise (1) and (2) if needed.
Write conservation of mass (or weight) for each component and for the entire system. Modify the diagram as new information is uncovered.
Solve for the unknowns.
Reflect on your solution. Do the concentrations or quantities make sense?
Avoid trying to just solve these problems in your head. Use the systematic approach above.

8. Class Problem: A 245 quart aquarium contains 3% salt by weight
Class Problem: A 245 quart aquarium contains 3% salt by weight. How much 10% salt by weight water would you need to add to bring the salt concentration to 4.5% salt by weight?
Useful Conversions: 1L = 0.001m3 = quarts = 0.264gal = 61.02in3
𝑋 kg salt water
10% NaCl
90% H2O
Convert quarts into mass:
245𝑞𝑢𝑎𝑟𝑡𝑠∙ 1𝐿 𝑞𝑢𝑎𝑟𝑡 ∙ 1𝑘𝑔 1𝐿 =231.85𝑘𝑔
kg salt water
245 quarts salt water
𝑌 kg salt water
3% NaCl
4.5% NaCl
97% H2O
95.5% H2O
Apply Conservation of Mass Equations:
𝑚 𝑖𝑛 − 𝑚 𝑜𝑢𝑡 + 𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 − 𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 = 𝑚 𝑓𝑖𝑛𝑎𝑙 − 𝑚 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
overall:
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
salt:
𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡
water:
231.85kg
+ 𝑋 =𝑌
0.03(231.85𝑘𝑔)
+ 0.1𝑋
=0.045∙𝑌
𝑘𝑔 +0.9X
=0.955∙𝑌
(1)
(2)
(3)
From (3):
𝑌= 𝑘𝑔 +0.9X
Plug this into (1):
𝑋= 𝑘𝑔 +0.9X −231.85𝑘𝑔
𝑋=235.49𝑘𝑔 𝑋−231.85𝑘𝑔
0.0576𝑋=3.641𝑘𝑔
𝑋=63.23 𝑘𝑔
Water with 10% salt by weight