Measuring Mass - Examples

1.50 grams of copper sulfate is needed to prepare a solution. Using a balance that reads to the hundredths place (2 decimals), this could be done as follows:

Direct Weighing - while this works, remember, we never weigh chemicals directly on the balance pan

1.	The balance is zeroed with nothing on the pan.
2.	Copper sulfate is added to the balance pan until the balance reads 1.50 g.

Weighing by Difference - Method 1

1.	The balance is zeroed with nothing on the pan. The balance should read 0.00.
2.	A folded piece of weighing paper is placed on the balance pan and the mass recorded. This piece weighs 0.38 g.
3.	Copper sulfate is added to the weighing paper until the balance reads 1.88 g. (0.38 g + 1.50 g = 1.88 g).
4.	The mass of copper sulfate is calculated by subtracting the first reading (step #2) from the second reading (step #3). For this example, 1.88 g - 0.38 g = 1.50 g.

Weighing by Difference - Method 2

1.	The balance is zeroed with nothing on the pan. The balance should read 0.00.
2.	A container of copper sulfate is placed on the balance pan and the mass recorded. This container plus the copper sulfate it contains weighs 31.56 g.
3.	Copper sulfate is removed from the container until the balance reads 30.06 g. (31.56 g - 1.50 g = 30.06 g).
4.	The mass of copper sulfate removed from the container is calculated by subtracting the second reading (step #3) from the first reading (step #2). For this example, 31.56 g - 30.06 g = 1.50 g.

By Taring the Balance

1.	The balance is zeroed with nothing on the pan. The balance should read 0.00.
2.	A folded piece of weighing paper is placed on the balance pan and the tare button pushed. The balance should read 0.00.
3.	Copper sulfate is added to the balance pan until the balance reads 1.50 g.
4.	Since the balance has already subtracted the mass of the weighing paper, the mass of copper sulfate is equal to the balance reading (1.50 g).