Chem HW - CALCULATIONS OF COPPER CYCLE LABORATORY¶
The Copper Cycle Diagram: One way to summarize the Copper Cycle Laboratory is to follow the procedure as an actual cycle using this graphic. Review this cycle and complete the table below.¶
I decided to show the work in the table so that it makes more sense logically.
| Cycle Step | Chemical Name | Chemical Formula | Element or Compound? | Mass | Molar Mass | Calculated Moles | Calculated Number of Particles |
|---|---|---|---|---|---|---|---|
| Initially, you measured 0.10 g of copper powder. | copper | Cu | Element | 0.10g | \(\(Cu(63.546AMU) = 63.546 \frac{g}{mol}\)\) | \(\(\frac{0.10 g}{63.546 \frac{g}{mol}} = 0.00157 mol\)\) | \(\(0.00157 mol * 6.02* 10^{23} = 9.4514 * 10^{20}\)\) |
| In step 2, 0.25 g of copper (II) hydroxide was made. | copper (II) hydroxide | Cu(OH)2 | Compound | 0.25g | \(\(Cu(63.546AMU) + 2O(15.999AMU) + 2H(1.008) = 97.561 \frac{g}{mol}\)\) | \(\(\frac{0.25 g}{97.561 \frac{g}{mol}} = 0.00256 mol\)\) | \(\(0.00256 mol * 6.02* 10^{23} = 1.54122 * 10^{21}\)\) |
| In step 3, 0.20 g of a black powder called copper (II) oxide was made. | copper (II) oxide | CuO | Compound | 0.20g | \(\(Cu(63.546AMU) + O(15.999AMU) + 79.545 \frac{g}{mol}\)\) | \(\(\frac{0.20 g}{79.545 \frac{g}{mol}} = 0.00251 mol\)\) | \(\(0.00251 mol * 6.02* 10^{23} = 1.51102 * 10^{21}\)\) |
| In step 5, you added 0.10 g Zn filings. | zinc | Zn | Element | 0.10g | \(\(Zn(65.38AMU) = 65.38 \frac{g}{mol}\)\) | \(\(\frac{0.10 g}{65.38 \frac{g}{mol}} = 0.00152 mol\)\) | \(\(0.00152 mol * 6.02* 10^{23} = 9.1504 * 10^{20}\)\) |
Last update:
June 5, 2023
Created: June 5, 2023
Created: June 5, 2023